Conference rusure::math

Gee, I got 100% on my abstract algebra final exam 20 years ago in college,
but would you believe I've FORGOTTEN what a group is ?

Please correct my fog.  What I recall is that for a group of order 120, that
means we start with a set of 120 elements.

We make a chart, listing the elements across the top and down the side:


+  A B C D ... 120 elements

A
B
C
D
... same 120 elements

The + represents a binary operation, not necessarily arithmetic plus.  The
"binary" means it takes 2 values.  For example, if

	A + C = B

then we start filling in the chart like this:

+  A B C D ... 120 elements

A      B
B
C
D
... same 120 elements


So, when you ask how many groups of order 120 there are, I guess you're asking
how many ways there are of filling in the chart.

But this is where my mind gets foggy.  Isn't it true that in order for it to
be a group, there are restrictions on how we define +, in other words we
can't just fill in the chart any old way.

My recollection is that + needn't be commutative.  In other words, if

	A + C = B

then it isn't necessarily so that

        C + A = B

(Now you KNOW we're not talking arithmetic + here!  Wouldn't the banks love
this one...)

Why is that my recollection ?  Because I recall that the term

	abelian group

means a group that IS commutative.  If we need an adjective to describe
a commutative group, then I assume there are groups that aren't.

o.k. what else do I recall about the restrictions of + for a group.  I believe
CLOSURE is a requirement.  That merely means that every element of the chart
will be filled in by one of the 120 elements themselves.

So, for example, if the 120 elements are the integers 1 through 120, then
we can't use arithmetic division, or even typical computer integer division for that
matter, as our operator, since 3 / 4 is 0 in computer integer division, and 0
isn't in the set of integers from 1 to 120.

Are there more restrictions ?

If not, then it seems to me that each of the 120*120 elements of the chart
may be filled in with one of 120 possibilities, and hence the total number
of groups would be

	120 * 120 * 120 ... there's 120^2 of these !

But you mentioned isomorphism.  What do I remember about that ?

I think that's when you take one group (i.e. one entirely filled in chart)
and replace each element by a different one.  For example, if the 120 elements
were the numbers 1 through 120, and we filled in the chart and then exchanged
all the 3's and 4's (on the border too, or just in the middle ???) then we'd
have an isomorphism.



Does this help ?

/Eric

    Re .1
    
    	I don't think there is an exposition of Group Theory in this
    conference. The best way to bring yourself back up to speed would
    be to read Gorenstein's superb exposition of group theory in Scientific
    American, referenced in 390.0. This takes someone from a state of
    complete innocence (or amnesia in your case, Eric :-) to an
    understanding the notion of the massive Classification Theorem for Finite
    Simple Groups.
    
    Cheers,
    Andy.

        I count three Abelian groups of order 120.  Still working on
        the non-Abelian groups of order 120.  I quickly found five.
        GAP could probably do this quickly.
        
        Dan

re .1

It was somewhat less than 20 years ago for me but I have still forgotten
most of it too.

My recollection is that a group (a set and an operation) requires closure,
an identity element and for each member of the set to have an inverse.
Consequently the number of groups is much less than the combinatorial limit.

Its been a while since I've done this stuff, but to may recollection, the
way approach this is look at the kernels and factor groups.  If NG(n) is the
number of groups of order n, then:

		NG(120) = NG(8*3*5)
			= NG(8)*NG(3)*NG(5)

Since 3 and 5 are primes, the only groups of order 3 and 5 are the cyclic
groups of those orders (call them C(3) and C(5) respectively), so NG(3) = NG(5)
= 1.  To the best of my recollection, there are 3 groups of order 8: C(8),
C(2)xC(4) and C(2)xQ, where Q is the quaterian group.  So overall, there are
three groups of order 120.

    Re .5:
    
    Your analysis is valid for commutative groups.  Non-commutative groups
    generally cannot be broken down so easily.
    
    
    				-- edp
    
    
Public key fingerprint:  8e ad 63 61 ba 0c 26 86  32 0a 7d 28 db e7 6f 75
To find PGP, read note 2688.4 in Humane::IBMPC_Shareware.

Re .5 and .6:

My memory is rusty, and I made the problem simpler than it really is, but the
basic approach I think is correct.  As indicated, you need to look at how 120
breaks down into factors.  For example, 120 = 2*60.  There is only one group of
order 2 and only 1 simple group of order 60.  Then you can look at how groups
of order 60 factor and so on.  

	There are two kinds of group: solvable (case S) and insolvable (case I).

	The only non-Abelian simple group of order < 168 is A5 (order 60).
Therefore, an insolvable group of order 120 must have A5 in its decomposition
series. There are two possible series:

(case I1)	I <| H <| G, where H == A5, and G/H == C2.

(case I2)	I <| H <| G, where H == C2, and G/H == A5.



	In this reply, we treat case (I1).

	Let x lie in G\A5. Conjugation by x acts as an automorphism of A5. What
is the (outer) AM group of A5, Out(A5)? Any relabeling of the 5 elements
in {1,2,3,4,5} evidently induces an AM. So [a group IM to] S5 lies in Out(A5). 
Are there any other elements in Out(A5)? A5 is generated by (123) & (345).
Suppose that (123) -> (abc) & (345) -> (def) under some AM �. a,b & c
must be disjoint, as must d, e and f. What is z = |{a,b,c}^{d,e,f}|?
	z = 1:
	wlog: c = d, and � maps (1,2,3,4,5) -> (a,b,c,e,f). So � is just one
of the AMs we already have.
	z = 2:
	(abc)(def) = (ghi) or (gh)(ij) for some g,h,i,j. But we know that
(123)(345) = (12345). Contradiction.
	z = 3:
	(abc) = (def) or (def)^-1. But (123) /= (345) or (345)^-1. 
Contradiction.
	So Out(A5) == S5.

If conjugation by x induces an AM �, then conjugation by xn (where n lies in A5)
induces an an AM �� (where � is the Inner AM induced by conjugation with n.)
Inn(A5) == A5. So in order to find the set of possible groups G, we need only 
consider two values of �, say I & (12).

We need also to determine x�, which must lie in A5. Since conjugation by x�
must induce the AM ߲, we know that x� is the identity.

So, there are exactly two groups, which we can recognize as:
	A5 x C2
	S5

Note: the general form of the attack we used here is one which we can use for 
the solvable groups (case S):
	Find a normal subgroup, N, of prime index, p.
	Postulate an element x lying in G\N
	Identify the possible sutomorphisms which conjugation by x can induce
	Remove redundant automorphisms by examining conjugagion by xn, n in N
	Establish the possible values of x^p, consistent with �^p (known)
	[If necessary remove redundant x^p by appeal to symmetry in N, etc.]
	Try to enumerate known groups, and match them to results of the above.

Note: this approach will not help us to deal with solvable groups where A5
appears as a factor group (i.e. case I2).

Let |G|=120.  By Sylow, the number of Sylow p-groups is == 1 (mod p) and divides
120/p.  So,

  there are 1,3,5, or 15 Sylow 2-groups (of order 8),
  there are 1,4,10, or 40 Sylow 3-groups (of order 3),
  there are 1 or 6 Sylow 5-groups (of order 5).

Clearly a Sylow 3- or 5-group is cyclic; and there are just 4 groups of order 8,
namely Z2^3, Z2xZ4, Z8, and D4.  Proof: if there is an element of order 8, the
group is clearly Z8.  If all elements /= e are of order 2, the group is Abelian
(an old result of Burnside: abab = e = aabb, so ba = ab) and hence is Z2^3; the
remaining case is that there is an element x of order 4 but none of order 8.  If
Abelian, we have Z2xZ4, otherwise an element y not a power of x must induce the
only nontrivial automorphism of Z4 = <x> under conjugation, so yxy' = x' (using
' for inverses).  Then yy must commute with x, so it must be in Z4; it can't be
x or x', so yy = xx.  The multiplication table can now be completed, and we must
have D4.

Also, no 8-group has an automorphism of order 5: if it did, consider the
permutation induced by conjugation by that element.  With 8 letters being
permuted, the result would have to be a 5-cycle fixing 3 elements.  One of these
3 is clearly e; call the other two a and b.  ab is then also fixed, and since it
isn't = a or b it must be e, so b=a'.  But this means a' /= a, so a must be of
order 4 or 8, which would imply that at least 4 (resp. 8) elements would be
fixed, a contradiction.

The implication of this is that any 5-element that normalizes a Sylow 2-group
here must centralize it.  Similarly, but more easily, any 3-element that
normalizes a 5-group must centralize it, and vice versa.

We also have, for q=3 or 5, that any 2 distinct q-groups intersect only in {e},
so the total number of q-elements is (q-1) times the number of (Sylow) q-groups.

If, then, a Sylow 5-group H5 is normal, there is a 3-element in its normalizer;
all 3-elements are conjugates of either this element or its square, so all
3-elements normalize H5, so all 3-elements centralize it, so all 5-elements
centralize all 3-elements, so 5 divides |N(H3)| for any H3, so the number of
Sylow 3-groups is either 1 or 4.  Similarly, if there is just 1 Sylow 3-group,
the number of Sylow 5-groups is 1.  That is:

  1 Sylow 3-group => 1 Sylow 5-group
  1 Sylow 5-group => 1 or 4 Sylow 3-groups.

Similarly, 1 Sylow 2-group => 1 Sylow 5-group (the 5-elements must centralize
the 2-group in this case, since they normalize it).

Therefore, if *any* Sylow group is normal, the Sylow 5-group is; and otherwise
there are exactly 6 Sylow 5-groups.

Re: -.1

>Clearly a Sylow 3- or 5-group is cyclic; and there are just 4 groups of order 
>8, namely C2^3, C2.C4, C8, and D4.  
    
    There's also Q, the quaternion group, comprising {�1,�i,�j,�k} where
    i�=j�=k�=-1, ij=k, ji=-k, etc. However, this doesn't affect your
    analysis of G.

>Then yy must commute with x, so it must be in C4; 
    
    C4 has index 2, so y� must be in C4. Same conclusion.
    
>it can't be x or x', so yy = xx. The multiplication table can now be completed
>and we must have D4.
    
    (1) If y� = x� as suggested, then the group is Q.
    (2) If y� = 1, then the group is D4.
    
>Also, no 8-group has an automorphism of order 5.
    
    Agreed. Let Z(G'), Inn(G') & Aut(G') be the centre, inner AM group and
    outer AM group of some group G'. Inn(G') == G'/Z(G'). Inn(G') <| Aut(G')
    
    G'		C2�	C2.C4	C8	D4	Q
    ----------------------------------------------
    Z(G')	C2�	C2.C4	C8	C2	C2
    Inn(G')	I	I	I	C2�	C2�
    Aut(G')    *GL3(2)	D4	C2�	D4	S4
     
    (*) This group has order 168, and is incidentally simple.

>if it did, consider the permutation induced by conjugation by that element.  

    An AM is not necessarily an inner one. However, this does not affect your 
proof, which hinges on the idea that the set of fix-points of an AM of G' forms 
a subgroup of G'.

    I agree with the rest of your reply.

    Note that if H3 is normal, then there is a subgroup of order 60 in G. (Such
a group is obviously normal in G, since it's of index 2.)

Proof: Suppose H3 is normal. 

        (1) The composition series for G contains C3 as a factor. If G were
insoluble, then its composition series would be {C2, A5} in some order. By the
Jordan-H�lder theorem, valid composition series must be anagrams of one
another. Contradiction. So G is soluble.

        (2) Phillip Hall's theorem for soluble groups states that if |G| = nm,
where (n,m) = 1, then any subgroup of order dividing n is contained in a 
subgroup of order n. (Additionally, all subgroups of order n are conjugate in 
G.) So, our G has a subgroup, K, of order 40.

        (3) By the usual Sylow counting, K has a normal Sylow-5-subgroup, K5. 
Again by Sylow, there will be a subroup of order 4, K4. Since K5 <| K, the
set K4.K5 constitutes a group, K20, of order |K4|.|K5|/|K4^K5| = 20. "^"
denotes intersection here. (One of the IM theorems.) So K has a subgroup of 
order 20.

        (4) So our G has a subgroup, H20, of order 20. By the same IM theorem,
since H3 <| G, H20.H3 constitutes a subgroup of order 60.

        If we have a normal H5, then the same result *nearly* holds. However
step 3 breaks down. We find ourselves examining a group, L, of order 24, and it
is not the case that there will always be a subgroup of order 12.

        If L lacks a subgroup of order 12, L is still soluble, so it must have
a normal subgroup of index 3. This is the Sylow-8-subgroup, L8. There must be
4 Sylow-3-subgroups. No we can repeat Mike's trick. Some 3-element must lie 
outside C(L8), the centralizer of L8, else L8 centralizes L3 which would imply
that L3 is normal in L. Contradiction. So L8 admits some non-trivial AM of 
order 3. So L8 is IM to C2� or Q. ...

        Before going down this path, I want to write some theory.

        Also, if G has a normal subgroup of order 4, 12, 15 or 20, then it has
a subgroup of order 60. Sketch of proof: We are looking for two subgroups, one 
normal in G, with trivial intersection, and orders multiplying to 60. By
Sylow/Hall, we know that subgroups of order 3, 4, 5, 15 exist.

        What's the point of all this? If we assume that G has a subgroup of
index 2, and if we can enumerate the 13 groups of order 60, then we can
compute a whole bunch of groups of order 120, in the same kind of way that Mike
produced the groups of order 8 from looking at the groups of order 4. 

        But we then need to look at all the groups which *don't* have a
subgroup of order 60. The more we can constrain these (e.g: cannot have normal
Sylow-3-subgroup) the better.

Cheers,
Andy.

[Note 1202.5 gives some theory. This reply covers the same basic ground, but
is substantially corrected & revised.]

	Given a finite group N and a prime p, what is a reasonable way to
enumerate all the groups G such that N is a normal subgroup of G, and [G:N] = p?

	Let Aut(N) be the group of automorphisms of N, let Inn(N) be the group 
of inner automorphisms of N (ie those which are induced by conjugation by some 
element of N). Let Z(N) be the centre of N. Let g' denote g^(-1). And let's 
remember that:
	Inn(N) is normal in Aut(N)
	Inn(N) is IM to N/Z(N)

	Pick � in Aut(N), � in N, such that:
		(1) �^p is the inner AM afforded by conjugation by �
		(2) � fixes �

	Then let P = {0,1,...p-1} have addition defined upon it, and let
'.' denote the group operation already defined on N, then we'll define an 
operation * on the set N x P, which we'll call G, such that;

	(a, i)*(b, j) = (a.�^i(b), i+j)			(if i+j < p)
		      = (a.�^i(b).�', i+j-p)		(if i+j >= p)

Theorem: G is a group under *, with a normal subgroup IM to N, of index p.

        The proof is tedious, and there are no surprises. Associativity is the 
trickiest to demonstrate. The identity is (1,0). (a,0) has inverse (a',0). (a,i)
has inverse (�^(-i)(a').�,p-i). Henceforth, we will blur the notation and regard
N as being a normal subgroup of G. A key element of G is (1,1), conjugation by
which induces the AM �.

	The question then is, how is it possible to simplify the search for
� & p, to avoid blind alleys and repetition?   Some immediate observations
relate the orders of N, Aut(N) & Z(N) the centre of N.   Label them n,a & z
respectively.   z|n & n|za, since Z(N) =< N & N/Z(N) == Inn(N) =< Aut(N).   
Let '$' denote 'does not divide'.

>	(i) Wlog, � can be chosen to have order a power of p, by consideration
>of the Sylow-p-subgroup of G.  So this gives us another constraint:
>		(3) � has order a power of p.
[Christmas! I haven't a clue how I proved this one. Is it even true? Let's 
forget it, and approach things from a different tack...]

	EITHER � lies in Inn(N) or it lies in Aut(N) \ Inn(N).

	(A) � lies in Inn(N).
	Conjugation by some element x in G\N affords �. What about conjugation
by some other element y = xu, u lying in N? Conjugation afforded by an element 
in xN will afford an element in �.Inn(N). Since � lies in Inn(N) we can find �
in N, such that conjugation by y = x� affords I. I is the inner AM afforded
by conjugation by any z in Z(N). A fortiori, I fixes z.

	So, wlog, if � lies in Inn(N), we can take �=I, � in Z(N), and p can
be anything. A choice remaining is which element of Z(N) to pick for �. Divide
Z(N) up into orbits under Aut(N), and only consider one representative from
each such orbit.

	(B) � lies in Aut(N)\Inn(N). As above, conjugation by some element in
xN affords an element in �.Inn(N). Take this conjugation to the pth power:
(xN)^p affords (�.Inn(N))^p. But Inn(N) <| Aut(N), so we can write that (xN)^p 
affords �^p.M, where M is a subset of Inn(N). Since �^p lies in Inn(N), we *may*
be able to find � in N such that conjugation by y = (x.�)^p affords I. But we
cannot be sure.

        (B1) One special case is where Inn(N) = I. I.e. N is abelian. In this
case, a fortiori, M = Inn(N). Here �^p = I, � can be any element in N fixed by
�, fix(�). Note: p|a. Once again, a choice remaining is which element to
pick for �. Divide fix(�) up into orbits under Aut(N), and only consider one
representative from each such orbit.

        (B2) If Inn(N) is non-trivial, then we are in a less controlled 
situation. We know that pn|za. Otherwise we know no more than the two criteria
given above relating � & �.

--------------------------------------------------------------------------------

        Here's a simple example to show these ideas at work.

What are the groups of order 12?

First, list them off the top of my head. (I think there are 5, I'm missing one.)

        1  2  3  4  6 12
C12     1  1  2  2  2  4
C6.C2   1  3  2     6
D6      1  7  2     2
A4      1  3  8

So let's find any missing ones. The possible subgroups of prime index are:


G'              C6      S3      C4      C2�

Z(G')           C6      I       C4      C2�
Inn(G')         I       S3      I       I
Aut(G')         C2      S3      C2      S3


(A) � = I. � lies in Z(G'). p can be anything. Representatives of conjugacy 
classes are:

                1       1       1       1
                a�              a�      a
                a�              a
                a

Each of these possibilities yields a group. If we take p = 2 or 3 as
appropriate, we get:

                C6.C2   S3.C2   C12     C6.C2
                C6.C2           C12     C6.C2
                C6.C2           C12
                C12

[There seems to be a lot of duplication here. Suggestions to improve?]

(B) For the 4 groups being considered above, if Aut(G') \ Inn(G') is non-empty, 
then Inn(G') = I. So we are looking just at case (B1).

Since p|a, C4 is not interesting to us, and we are looking at C6 (p=2) and
C2� (p=3).

C6. p=2, �=(a->a'). � is fixed by �, i.e.: 1 or a�. The groups derived are D6 
and X <--- the missing group.

        1  2  3  4  6 12
C12     1  1  2  2  2  4
C6.C2   1  3  2     6
D6      1  7  2     2
A4      1  3  8
X       1  1  2  6  2

The orders of the elements is different from any of the others, so we have
found a new group. A presentation for it is:

< a, b | a^6, b'aba, b�a� >

C2�. p=3, �=(a->b->c->a). � is fixed by �, i.e.: 1. The group derived is A4.

--------------------------------------------------------------------------------

That's it!
Andy

        Up to isomorphism there are 14 non-Abelian groups of order 120
        with at least one element of order 60 but no element of order
        greater than 60.
        
        	Note:  Actually, an element of order greater than
        	60 would have order 120.  That would make the group
        	isomorphic to the cyclic group C120, which is Abelian.
        
        So let G be a non-Abelian group of order o(G) = 120, let a be
        an element of G of order o(a) = 60, let H = <a> be the sixty
        element subgroup of G generated by a, and let b be an
        arbitrary element of G - H.
        
        [G:H] = |G|/|H| = 120/60 = 2.  All subgroups of index 2 are
        normal subgroups, so H is a normal subgroup of G.  The cosets
        are H and G-H, so for example bH = Hb = G-H.
        
        So we know that G = H union Hb
        = { a^i | 0 <= i < 60 } union { a^i b | 0 <= i < 60 }
        
        Using those "names", what is the multiplication table for G?
        Two kinds of muliplications of group elements are easy:
        
        	a^i a^j = a^(i+j mod 60)
        	a^i (a^j b) = a^(i+j mod 60) b
        
        The other two can be determined once ba and b^2 are known.
        Since ba is an element of bH = Hb, there must be an r with
        0 <= r < 60 such that ba = a^r b.  Then b(a^2) = b(aa) =
        (ba)a = (a^r b)a = (a^r)(ba) = (a^r)^2 = a^(2r mod 60).
        Likewise, ba^k = a^(kr mod 60).
        
        The coset Hb^2 is not equal to the coset Hb, so it must be
        equal to the coset H, which means there is an s with 0 <= s
        < 60 such that b^2 = a^s.
        
        Now we can enumerate all possibilities: for each r and s in
        the given ranges, fill in a potential multiplication table
        which using ba = (a^r)b, b^2 = a^s becomes
        
        	a^i a^j = a^(i+j mod 60)
        	a^i (a^j b) = a^(i+j mod 60) b
        	(a^i b) a^j = a^(i+rj mod 60) b
        	(a^i b) (a^j b) = a^(i+rj+s mod 60)
        
        and verify that the table table does indeed give a group
        (i.e., it is associative and a^0 is an identity element
        relative to which every element has an inverse), weed out the
        (r,s) that give Abelian groups, and check for isomorphisms
        among the rest.
        
        A little more analysis first.  Since ba = a^r b, it follows
        that ba(b^-1) = a^r.  Now ba(b^-1) is just conjugation of a
        by b^-1, i.e., the image of a under the inner automorphism
        og conjugation by b^-1.  Because that is an automorphism of G
        it must be the case that o(a^r) = o(ba(b^-1)) = o(a) = 60.
        Thus we can further limit r to 0 <= r < 60 and gcd(60,r) = 1.
        So r is one of
        
        	1 7 11 13 17 19 23 29 31 37 41 43 47 49 53 59
        
        and 0 <= s < 60.
        
        Furthermore, since G = <a,b> if we have r=1 then ba = ab and
        the group is Abelian.
        
        	Note:  In fact, r=1,odd s leads to C120 and r=1,even s
        	results in C2 x C60.
        
        So we need only check
        
        	r one of 7 11 13 17 19 23 29 31 37 41 43 47 49 53 59
        	0 <= s < 60
        
        Let a computer grind through the rest of the work.  Here is a
        summary of the 14 Non-Abelian 120-groups with an element of
        order 60.  The raw results are shown below.
        
        
          count of elements of order
  1  2  3  4  5  6  8 10 12 15 20 24 30 40 60120    r  s
 -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --   -- -------

  1  1  2  2  4  2 12  4  4  8  8  0  8 48 16  0 : 41  3mod6
  1  1  2  2  4  2 20  4  4  8  8 40  8  0 16  0 : 49  5mod10
  1  1  2  2  4  2 60  4  4  8  8  0  8  0 16  0 : 29 15mod30
  1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0 : 31  2mod4
  1  1  2 14  4  2  0  4  4  8 56  0  8  0 16  0 : 11  6mod12
  1  1  2 22  4  2  0  4 44  8  8  0  8  0 16  0 : 19 10mod20
  1  1  2 62  4  2  0  4  4  8  8  0  8  0 16  0 : 59 30
  1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0 : 31  0mod4
  1  7  2  8  4  2  0 28  4  8 32  0  8  0 16  0 : 41  0mod6
  1 11  2 12  4 22  0  4 24  8  8  0  8  0 16  0 : 49  0mod10
  1 13  2  2  4  2  0 52  4  8  8  0  8  0 16  0 : 11  0mod12
  1 21  2  2  4 42  0  4  4  8  8  0  8  0 16  0 : 19  0mod20
  1 31  2 32  4  2  0  4  4  8  8  0  8  0 16  0 : 29  0mod30
  1 61  2  2  4  2  0  4  4  8  8  0  8  0 16  0 : 59  0
        
        Dan
        
        
          count of elements of order
  1  2  3  4  5  6  8 10 12 15 20 24 30 40 60120    r  s     description
 -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --   -- -- -------------------
        
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1  0 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1  1 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1  2 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1  3 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1  4 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1  5 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1  6 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1  7 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1  8 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1  9 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 10 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 11 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 12 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 13 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 14 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 15 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 16 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 17 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 18 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 19 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 20 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 21 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 22 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 23 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 24 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 25 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 26 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 27 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 28 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 29 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 30 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 31 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 32 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 33 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 34 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 35 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 36 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 37 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 38 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 39 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 40 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 41 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 42 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 43 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 44 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 45 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 46 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 47 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 48 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 49 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 50 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 51 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 52 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 53 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 54 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 55 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 56 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 57 Abelian cyclic C120
  1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 :  1 58 Abelian C2 x C60
  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1 59 Abelian cyclic C120
  1 13  2  2  4  2  0 52  4  8  8  0  8  0 16  0 : 11  0 (11  0)
  1  1  2 14  4  2  0  4  4  8 56  0  8  0 16  0 : 11  6 (11  6)
  1 13  2  2  4  2  0 52  4  8  8  0  8  0 16  0 : 11 12 (11  0)
  1  1  2 14  4  2  0  4  4  8 56  0  8  0 16  0 : 11 18 (11  6)
  1 13  2  2  4  2  0 52  4  8  8  0  8  0 16  0 : 11 24 (11  0)
  1  1  2 14  4  2  0  4  4  8 56  0  8  0 16  0 : 11 30 (11  6)
  1 13  2  2  4  2  0 52  4  8  8  0  8  0 16  0 : 11 36 (11  0)
  1  1  2 14  4  2  0  4  4  8 56  0  8  0 16  0 : 11 42 (11  6)
  1 13  2  2  4  2  0 52  4  8  8  0  8  0 16  0 : 11 48 (11  0)
  1  1  2 14  4  2  0  4  4  8 56  0  8  0 16  0 : 11 54 (11  6)
  1 21  2  2  4 42  0  4  4  8  8  0  8  0 16  0 : 19  0 (19  0)
  1  1  2 22  4  2  0  4 44  8  8  0  8  0 16  0 : 19 10 (19 10)
  1 21  2  2  4 42  0  4  4  8  8  0  8  0 16  0 : 19 20 (19  0)
  1  1  2 22  4  2  0  4 44  8  8  0  8  0 16  0 : 19 30 (19 10)
  1 21  2  2  4 42  0  4  4  8  8  0  8  0 16  0 : 19 40 (19  0)
  1  1  2 22  4  2  0  4 44  8  8  0  8  0 16  0 : 19 50 (19 10)
  1 31  2 32  4  2  0  4  4  8  8  0  8  0 16  0 : 29  0 (29  0)
  1  1  2  2  4  2 60  4  4  8  8  0  8  0 16  0 : 29 15 (29 15)
  1 31  2 32  4  2  0  4  4  8  8  0  8  0 16  0 : 29 30 (29  0)
  1  1  2  2  4  2 60  4  4  8  8  0  8  0 16  0 : 29 45 (29 15)
  1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0 : 31  0 (31  0)
  1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0 : 31  2 (31  2)
  1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0 : 31  4 (31  0)
  1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0 : 31  6 (31  2)
  1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0 : 31  8 (31  0)
  1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0 : 31 10 (31  2)
  1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0 : 31 12 (31  0)
  1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0 : 31 14 (31  2)
  1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0 : 31 16 (31  0)
  1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0 : 31 18 (31  2)
  1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0 : 31 20 (31  0)
  1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0 : 31 22 (31  2)
  1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0 : 31 24 (31  0)
  1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0 : 31 26 (31  2)
  1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0 : 31 28 (31  0)
  1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0 : 31 30 (31  2)
  1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0 : 31 32 (31  0)
  1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0 : 31 34 (31  2)
  1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0 : 31 36 (31  0)
  1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0 : 31 38 (31  2)
  1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0 : 31 40 (31  0)
  1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0 : 31 42 (31  2)
  1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0 : 31 44 (31  0)
  1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0 : 31 46 (31  2)
  1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0 : 31 48 (31  0)
  1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0 : 31 50 (31  2)
  1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0 : 31 52 (31  0)
  1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0 : 31 54 (31  2)
  1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0 : 31 56 (31  0)
  1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0 : 31 58 (31  2)
  1  7  2  8  4  2  0 28  4  8 32  0  8  0 16  0 : 41  0 (41  0)
  1  1  2  2  4  2 12  4  4  8  8  0  8 48 16  0 : 41  3 (41  3)
  1  7  2  8  4  2  0 28  4  8 32  0  8  0 16  0 : 41  6 (41  0)
  1  1  2  2  4  2 12  4  4  8  8  0  8 48 16  0 : 41  9 (41  3)
  1  7  2  8  4  2  0 28  4  8 32  0  8  0 16  0 : 41 12 (41  0)
  1  1  2  2  4  2 12  4  4  8  8  0  8 48 16  0 : 41 15 (41  3)
  1  7  2  8  4  2  0 28  4  8 32  0  8  0 16  0 : 41 18 (41  0)
  1  1  2  2  4  2 12  4  4  8  8  0  8 48 16  0 : 41 21 (41  3)
  1  7  2  8  4  2  0 28  4  8 32  0  8  0 16  0 : 41 24 (41  0)
  1  1  2  2  4  2 12  4  4  8  8  0  8 48 16  0 : 41 27 (41  3)
  1  7  2  8  4  2  0 28  4  8 32  0  8  0 16  0 : 41 30 (41  0)
  1  1  2  2  4  2 12  4  4  8  8  0  8 48 16  0 : 41 33 (41  3)
  1  7  2  8  4  2  0 28  4  8 32  0  8  0 16  0 : 41 36 (41  0)
  1  1  2  2  4  2 12  4  4  8  8  0  8 48 16  0 : 41 39 (41  3)
  1  7  2  8  4  2  0 28  4  8 32  0  8  0 16  0 : 41 42 (41  0)
  1  1  2  2  4  2 12  4  4  8  8  0  8 48 16  0 : 41 45 (41  3)
  1  7  2  8  4  2  0 28  4  8 32  0  8  0 16  0 : 41 48 (41  0)
  1  1  2  2  4  2 12  4  4  8  8  0  8 48 16  0 : 41 51 (41  3)
  1  7  2  8  4  2  0 28  4  8 32  0  8  0 16  0 : 41 54 (41  0)
  1  1  2  2  4  2 12  4  4  8  8  0  8 48 16  0 : 41 57 (41  3)
  1 11  2 12  4 22  0  4 24  8  8  0  8  0 16  0 : 49  0 (49  0)
  1  1  2  2  4  2 20  4  4  8  8 40  8  0 16  0 : 49  5 (49  5)
  1 11  2 12  4 22  0  4 24  8  8  0  8  0 16  0 : 49 10 (49  0)
  1  1  2  2  4  2 20  4  4  8  8 40  8  0 16  0 : 49 15 (49  5)
  1 11  2 12  4 22  0  4 24  8  8  0  8  0 16  0 : 49 20 (49  0)
  1  1  2  2  4  2 20  4  4  8  8 40  8  0 16  0 : 49 25 (49  5)
  1 11  2 12  4 22  0  4 24  8  8  0  8  0 16  0 : 49 30 (49  0)
  1  1  2  2  4  2 20  4  4  8  8 40  8  0 16  0 : 49 35 (49  5)
  1 11  2 12  4 22  0  4 24  8  8  0  8  0 16  0 : 49 40 (49  0)
  1  1  2  2  4  2 20  4  4  8  8 40  8  0 16  0 : 49 45 (49  5)
  1 11  2 12  4 22  0  4 24  8  8  0  8  0 16  0 : 49 50 (49  0)
  1  1  2  2  4  2 20  4  4  8  8 40  8  0 16  0 : 49 55 (49  5)
  1 61  2  2  4  2  0  4  4  8  8  0  8  0 16  0 : 59  0 (59  0)
  1  1  2 62  4  2  0  4  4  8  8  0  8  0 16  0 : 59 30 (59 30)

        Now suppose G is a group of order o(G) = 120 and that G
        contains an element a of order o(a) = 40.  Let H = <a>
        be the subgroup (isomorphic to C40) of G generated by a.
        
        By the Sylow theorems, G contains a subgroup of order 3,
        and therefore an element b of order o(b) = 3.  It is easily
        shown that b is not in H, and that in fact
        
        	G = { a^i b^j | 0 <= i < 40, 0 <= j < 3 }
        
        We already know what (a^i1 b^j1)(a^i2 b^j2) is when either
        j1 = 0 or i2 = 0 (or both).  We can fill in the entire
        multiplication table of G once we also know both ba and b^2 a.
        For example, (a^5 b^2)(a^3 b) = a^5 (b^2 a) (a^2 b) and after
        substituting in b^2 a = a^x b^y we get (a^(5+x mod 40) b^y)(a^2 b).
        Note how the "i2" value has decreased by one; once we know
        
        	ba = a^r1 b^s1
        	b^2 a = a^r2 b^s2
        
        we can fill in the entire multiplication table and verify if
        it is indeed a group.
        
        There aren't that many possibilities, and computer search
        yields only two groups:
        
        	1. ba = ab, b^2a = ab^2	(this is C120, Abelian and cyclic)
        	2. ba = ab^2, b^2a = ab (non-Abelian)
        
          count of elements of order
  1  2  3  4  5  6  8 10 12 15 20 24 30 40 60120   r1 s1 r2 s2
 -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --   -- -- -- --

  1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 :  1  1  1  2 cyclic C120
  1  1  2  2  4  2 12  4  4  8  8  0  8 48 16  0 :  1  2  1  1
        
        The first is C120, the second has an element of order 60 and
        so must be one of the groups [it turns out to be the (41, 3mod6)
        group] from the previous reply.
        
        Dan

    Dan,
    
    	Neat work.
    
    	One point I'm troubled about. How do you tell whether two groups
    your program explores are the same? Do you just check whether the
    orders of the elements are the same? This isn't enough. For instance
    C4*C2 and the dihedral group of order 8 *both* have 1,4 & 2 elements of
    order 1,2 & 4 respectively.
    
    Cheers,
    Andrew.

>How do you tell whether two groups your program explores are the same?
        
        For the "C40 by 3" case, there was clearly an element of order
        60 and so the group had to be one of the "C60 by 2" groups.
        
        For the "C60 by 2" groups, with H = <a> = C60 and b in G - H,
        after determining that there was an (n,m) group [i.e., one
        with ba = a^n b, b^2 = a^m] the program checked all of the c
        in H with <c> = H and all of the d in G - H and found the
        n',m' such that dc = c^n' d and d^2 = c^m'.  So "the" (n,m)
        group [there is an isomorphism theorem in there] is also "the"
        (n',m') group.  I classified the group by the least m' (n' was
        always n) that specified it.
        
        Dan

    > n was always n'
    
    	Yes, this is correct, and is not just a question of chance.
    	Set c = a^p & d = ba^q, then it's easy to see that n = n', and also
    m' = [(n+1)q + m]P, where pP == 1 mod 60.
    
    Andrew.

Theorem: Up to IM, there are 3 insolvable groups of order 120.


Proof: An insolvable group is one which has a non-prime simple group in its
composition series. The only non-prime simple group of order < 168 is A5,
with order 60. Therefore, if G is an insolvable group of order 120, it
has composition series:

	[C2, A5]	(case I1)
or	[A5, C2]	(case I2)

case I1:
	As shown in 2005.???, there are just two groups up to IM:
		A5xS2
		S5

case I2:

	The concepts here are very simple, but it just takes a little time
to explain it all. Hidden in here is a beautiful idea: that a plane graph is
a union of two trees: one of vertices linked by edges, and the other of
faces linked by the remaining edges.

	Read note 1089.* to learn about Cayley & Schreier diagrams. [In 
fact, it was the specific "case I2" group which cropped up in that note which 
provoked 2005.0, a more general question about groups of order 120.] What 
is different here is that we *don't* start out with a presentation of the group.

	G has a normal subgroup F of order 2. G/F ~= A5. Draw a Cayley 
diagram for A5, with two generators: call them a & c, and the following
relators: a^5, (a.c)^3 & c^2. The diagram looks a bit like a soccer ball, with
12 pentagons, 20 hexagons, and 30 "2-sided ovals". Each pentagon is adjacent to
5 hexagons. Each hexagon is adjacent to 3 pentagons and 3 ovals. Each oval is
adjacent to 2 hexagons. Each vertex has indegree = outdegree = 2. I wish I 
could draw this pretty, highly symmetrical shape in Notes!
        
	Now let's abuse the same names, a & c, to refer to two elements of G
such that Fa & Fc map to a & c in the IM between G/F & A5. Say F = <f>. Since 
F is normal, any conjugate of f lies in F. But F only has 2 elements, and the 
identity is a conjugate only of itself. So f is central. This simplifies 
things considerably. It means that we don't need to consider how conjugation by
a or c permutes the elements of F. All we need to determine is whether each
of a^5, (a.c)^3 & c^2 equal 1 or f. Eight possible cases. 

	Let's regard the Cayley diagram for A5 as a Schreier diagram for G mod
F. Then what's neat is that this diagram is valid for exactly the eight
possible cases that we are trying to investigate!

	To do this, we "up the magnification" on the Schreier diagram, S, for 
G mod F, to turn it into a Cayley diagram, C, for G. Each vertex in S resolves
into a pair of vertices in C. For each vertex in S, arbitrarily prefer one of 
the two vertices in C: call it x and call its partner bx. An edge labelled a 
linking vertex Bx to vertex Bz, in S corresponds, in C, to a "twisted pair" of 
edges again labelled a, which:

	- EITHER map x to z and bx to bz (i.e. xa = z, call this valency 0)
	- OR map x to bz and bx to z (i.e. xa = bz, call this valency 1)

Similarly for the edges labelled c. Now what we are interested in is what 
happens when we "go round" a polygon (pentagon, hexagon or oval) in C, starting
from a vertex x. Do we return to the starting point x, or do we return to the 
other vertex bx. This depends on whether the sum mod 2 of all the valencies of 
the edges we transit is 0 or 1. Say that the desired valency of pentagons
= 0 or 1 if a^5 = 1 or f respectively. Similarly hexagons & (a.c)^3. Similarly
ovals & c^2.

	This is reminiscent of a Mobius strip: after going for a walk and 
coming back to the starting place, are we on the same side of the paper or the
other side?

	First, let's remove unnecessary degrees of freedom. Let T be a spanning
tree of S. Pick a vertex, and prefer one of the two corresponding vertices in C.
Now choose the preferences for all the vertices of S, such that each edge
in T has valency 0. 

	Consider an edge, e, in S\T. Together with certain edges from T, it
forms a cycle which encloses certain polygons. Define the valency of e to be 
simply the sum mod 2 of the desired valencies of the enclosed polygons.

	Note that since S appears on the surface of a sphere, there are
in fact *two* sets of enclosed polygons: the inside and the outside. Since
the sum mod 2 of the desired valency of all the polygons is 0 (there is an
even number of each type of polygon), the two sets give consistent
recommendations for the labelling of e.

	Now what is the actual sum mod 2 of the valencies of the edges
round each polygon? Firstly, the desired polygon valencies do constitute
a solution to this which is consistent by construction with the valencies
of the edges. Secondly, there are no other solutions. We have 62 polygons
which are linked by 61 non-tree edges. Let D be the graph with a vertex
corresponding to each polygon, and an edge corresponding to each edge in S\T.
D is connected, so is a tree. So D has a vertex, v, with degree 1. Call the
edge, e(v). The valency of (the polygon) v is determined by the valency of 
e(v). Prune v from D. D\v is still a tree, so iterate to find the valencies
of all the polygons. In the end, there is one polygon left. The valency of
that is chosen to ensure that the sum mod 2 of all the polygon valencies is 0 
mod 2.

	What all this means is that *each* of the 8 candidate groups is a real 
group of order 120 and normal subgroup order 2:

	G(i,j,k) = <f, a, c| f^2, [f,a], [f,c], a^5.f^i, (a.c)^3.f^j, c^2.f^k> 

where i,j,k drawn from {0,1}.

	Note that G(1,1,1) is IM to A5xC2, which we have already encountered,
in case I1. Now, are there any duplicates amongst the 8? By applying the
transformations a->fa and c->fc independently, we can see that G(1,1,k)
is IM to G(0,1,k), G(1,0,k) and G(0,0,k). However, in G(0,0,0), the Sylow-2-
subgroup is IM to C2^3, while G(0,0,1) contains an element of order 4.
So these 2 groups are not IM.

	re .10
        
>        What's the point of all this? If we assume that G has a subgroup of
>index 2, and if we can enumerate the 13 groups of order 60, then we can
>compute a whole bunch of groups of order 120, in the same kind of way that Mike
>produced the groups of order 8 from looking at the groups of order 4. 
>
>        But we then need to look at all the groups which *don't* have a
>subgroup of order 60. The more we can constrain these (e.g: cannot have normal
>Sylow-3-subgroup) the better.
        
        For a 120 element group, n5 is 1 or 6.  If n5 is 1, then there
        is a normal subgroup of order 40 or 60.  If n5 is 6, then the
        group is isomorphic to S5 or has a normal subgroup of order 2
        with factor group isomorphic to A5 or has a normal subgroup
        isomorphic to A5.  So I think you've already enumerated all of
        the 120-element groups with n5 = 6.
        
        Proof:  Use the following:
        
        	If H and K are both subgroups of a group G and one
        	of them is a normal subgroup of G, then <H,K> (i.e.,
        	the subgroup generated by their union) = HK = KH.
        
        	If H is a subgroup of a group G and A is the set of
        	right cosets of H and B is the set of conjugates of H
        	then there is a homomorphism from G onto a subgroup of
        	Sym(A) that acts transitively on A (and the kernel of
        	which is Core(H)) and there is homomorphism from G
        	onto a subgroup of Sym(B) which acts transitively on B.
        
        	If T is a homomorphism of a group G onto a group H and
        	U is a homomorphism of H onto a group K then the
        	composition T then U is a homomorphism of G onto K.
        
        	Enumerate the transitive subgroups of S6 (or at least
        	those with orders a factor of 120) and of S3.
        
        So if n5 = 1, then G has a normal subgroup H of order 5.  By
        the Sylow theorems it also has a subgroup K of order 8.  Then
        <H,K> = HK = KH is a subgroup of G of order 40.  So there is a
        homomorphism of G onto a transitive subgroup L of S3.  L
        either has order 3 (and so HK is a normal subgroup of G) or L
        is all of S3 and has a normal subgroup of index 2 (and so G
        also has a normal subgroup of index 2).
        
        If instead n5 = 6 then there is a homomorphism of G onto a
        transitive subgroup L of S6, where the order of L is a factor
        of 120.  Such a subgroup has size 6,12,24,60 (and is
        isomorphic to A5) or 120 (and is isomorphic to S5).  But
        wait, if the order of L were 24 then the kernel of the
        homomorphism from G onto L would have order 5, making n5 = 1.
        If the order of L is 6 then L has a normal subgroup of index
        2 and so G does as well.  All groups of order 12 have a normal
        subgroup of index 2 or 3 so if the order of L is 12 then G has
        a normal subgroup of index 2 or 3.  So either L is isomorphic
        to A5 or L and G are isomorphic to S5 or G has a normal
        subgroup H of index 2 or 3.  [G:H] can't be 3 because then H
        has order 40 with n5(H) = 1 but n5(H) = n5(G) = 6.  If [G:H]
        is 2 then H has order 60 with n5(H) = n5(G) = 6 and so H is
        isomorphic to A5.  So G is isomorphic to S5 or has either a
        normal subgroup or a factor group isomorphic to A5.
        
        Why is a 60-element group H with n5(H) = 6 isomorphic to A5?
        The homomorphism of H onto a transitive subgroup of S6 must
        have an image of size 6, 12, [oops, no can do, that implies a
        normal kernel of order 5 and so n5(H) = 1] or 60 (isomorphic
        to A5).  If 6, then H has a normal subgroup K or order 30 with
        n5(K) = 6 and that's impossible.  So the homomorphism is an
        isomorphism of H onto a subgroup of S6 isomorphic to A5.
        
        There are 14 isomorphism classes of 40-element groups and 13
        isomorphism classes of 60-element groups.  Proof:  point your
        web browser to
        
        	http://www.research.att.com/~njas/sequences/index.html
        
        and enter as an example sequence 1,1,1,2,1,2,1,5,2,2,1 and it
        will report back:

Matches (up to a limit of 10) found for  1 1 1 2 1 2 1 5 2 2 1  :

----------------------

%I A000001 M0098 N0035
%S A000001 1,1,1,2,1,2,1,5,2,2,1,5,1,2,1,14,1,5,1,5,2,2,1,15,2,2,5,4,1,4,1,51,
        1,2,
%T A000001 1,14,1,2,2,14,1,6,1,4,2,2,1,52,2,5,1,5,1,15,2,13,2,2,1,13,1,2,4,267,
        1,4
%N A000001 Number of groups of order n.
%R A000001 AJM 52 617 30. HS64. CM84 134.
%O A000001 1,4
%K A000001 nonn
%A A000001 njas
        
        Then track over to the entries for n=40 and n=60. :-)
        
        Dan

        Reply .17 enumerated the 3 isomorphism classes of 120-element
        groups in which the six 5-element subgroups are not normal.
        
        The alternative is that a 120-element group has a single
        5-element subgroup which is normal.  My program enumerated 44
        isomorphism classes of such groups.  [See below; all have
        exactly four elements of order five, as expected.]
        
        That should make 47 isomorphism classes of 120-element groups
        overall.
        
        Dan
        
        
        count of elements of order n
        in 120-element groups with
        one Sylow 5-subgroup
        
1  2  3  4  5  6  8 10 12 15 20 24 30 40 60 120
- -- -- -- -- -- -- -- -- -- -- -- -- -- -- ---
1  1  2  2  4  2  4  4  4  8  8  8  8 16 16 32 Abelian
1  1  2  2  4  2 12  4  4  8  8  0  8 48 16  0
1  1  2  2  4  2 20  4  4  8  8 40  8  0 16  0
1  1  2  2  4  2 60  4  4  8  8  0  8  0 16  0
1  1  2  6  4  2  0  4 12  8 24  0  8  0 48  0
1  1  2 10  4  2 20  4 20  8  0 40  8  0  0  0
1  1  2 10  4  2 60  4 20  8  0  0  8  0  0  0
1  1  2 14  4  2  0  4  4  8 56  0  8  0 16  0
1  1  2 22  4  2  0  4 44  8  8  0  8  0 16  0
1  1  2 46  4  2  0  4 20  8 24  0  8  0  0  0
1  1  2 62  4  2  0  4  4  8  8  0  8  0 16  0
1  1  8  6  4  8  0  4  0 32 24  0 32  0  0  0
1  3  2  4  4  6  0 12  8  8 16  0 24  0 32  0 Abelian
1  3  2 12  4  6  0 12  0  8 48  0 24  0  0  0
1  3  2 20  4  6  0 12 40  8  0  0 24  0  0  0
1  3  2 60  4  6  0 12  0  8  0  0 24  0  0  0
1  5  2  2  4 10  0 20  4  8  8  0 40  0 16  0
1  7  2  0  4 14  0 28  0  8  0  0 56  0  0  0 Abelian
1  7  2  8  4  2  0 28  4  8 32  0  8  0 16  0
1  7  2 40  4  2  0 28 20  8  0  0  8  0  0  0
1  7  8  0  4  8  0 28  0 32  0  0 32  0  0  0
1  9  2  6  4  6  0 36  0  8 24  0 24  0  0  0
1  9  8  6  4  0  0 36  0 32 24  0  0  0  0  0
1 11  2 12  4 22  0  4 24  8  8  0  8  0 16  0
1 11  2 20  4 22  0  4 40  8  0  0  8  0  0  0
1 11  2 36  4 22  0  4  0  8 24  0  8  0  0  0
1 11  2 60  4 22  0  4  0  8  0  0  8  0  0  0
1 13  2  2  4  2  0 52  4  8  8  0  8  0 16  0
1 13  2 10  4 26  0 12 20  8  0  0 24  0  0  0
1 15  2  0  4  6  0 60  0  8  0  0 24  0  0  0
1 17  2 30  4 22  0 28  0  8  0  0  8  0  0  0
1 21  2  2  4 42  0  4  4  8  8  0  8  0 16  0
1 23  2  0  4 46  0 12  0  8  0  0 24  0  0  0
1 23  2 40  4 10  0 12 20  8  0  0  0  0  0  0
1 23  8  0  4 40  0 12  0 32  0  0  0  0  0  0
1 31  2 16  4  2  0  4 20  8 24  0  8  0  0  0
1 31  2 32  4  2  0  4  4  8  8  0  8  0 16  0
1 33  2 30  4  6  0 12  0  8  0  0 24  0  0  0
1 33  8 30  4  0  0 12  0 32  0  0  0  0  0  0
1 37  2 10  4  2  0 28 20  8  0  0  8  0  0  0
1 41  2  6  4 22  0  4  0  8 24  0  8  0  0  0
1 47  2  0  4 22  0 28  0  8  0  0  8  0  0  0
1 61  2  2  4  2  0  4  4  8  8  0  8  0 16  0
1 63  2  0  4  6  0 12  0  8  0  0 24  0  0  0