Allen Knutson's Math 113, spring '04, HW #8, due April 1st (really!)

Let G be a group of size n, p a prime, such that |Syl_p(G)| = m > 1. If n doesn't divide m!, show that G has a nontrivial normal subgroup (neither {1} nor G). [A harder question, that I'm not asking: show the same for n not dividing m!/2. This is relevant for n=112.]

Let p^1 || |G|, and |Syl_p(G)| = m. Show that there are exactly m(p-1) elements of order p in G.

Let G be a group of order p^k, so Sylow's theorems become useless. If k>1, show that G has a nontrivial normal subgroup, i.e. isn't "simple".

Let G be a group of size n, n at most 100. Show that unless n is prime or 60, that G is not a simple group. If you can't handle all the cases, give a list of the cases that you get stuck on.

A1. Let G act on the Syl_p(G) (by conjugating them around), so we have a homomorphism phi : G -> permutations of Syl_p(G). This permutation group has m! elements, a hint we're on the right track.

Whenever we see ANY group homomorphism, we should look at what the First Isomorphism Theorem says:

ker phi --> G -->> G/ker phi --iso--> Im phi --inclusion--> perms We want to show that ker phi is nontrivial. If it's {1}, then G, G/ker phi, and Im phi all have |G| elements. But then we'd have a subgroup with size |G| of a group with size m!, contradicting Lagrange's theorem.

If it's all of G, then G/ker phi has one element, and every element of G is leaving every p-Sylow in place. This contradicts Sylow's second theorem, which says that G acts transitively. (Here we use the m>1 assumption -- otherwise leaving things, I mean thing, in place would be just fine.)

A2. Let S,T be two distinct Sylow p-groups. Each is of size actually p (not p^2, p^3, etc.) by the p^1 || |G| assumption. Their intersection is a subgroup of both, hence has size dividing p, hence is either of size 1 or p. But it's not p, since we assumed S,T distinct.

So they don't share elements other than the identity. Hence each accounts for exactly p-1 elements of order p, and there is no overcount. The total is m(p-1).

A3. We already proved that G has a nontrivial center Z(G). Let g be a nonidentity element of Z(G), of order p^j (since it has to divide p^k). Since it's not the identity, j isn't 0, so it's at least 1. Let h = g ^ {p^{j-1}}. Then h is of order p, and still in the center. Hence the cyclic group generated by h is of order p, and normal (check that central implies normal). Also, it's not the whole group, since k>1.

A4. First pass: if n = m p^k, for p>m (so p is the largest prime in n), then there is necessarily exactly one p-Sylow, hence normal. That kills an enormous number of cases. If m=1 and k=1, we get to ignore it. If m=1 and k>1, we can use A3 above.

The remaining numbers: 12, 24, 30, 36, 40 (which has only one 5-Sylow anyway), 45 (one 5-Sylow), 48, 56, 60 (which we're ignoring), 63 (one 7-Sylow), 70 (one 7-Sylow), 72, 80, 84 (one 7-Sylow), 90, 96, 100.

Another sweep: if p^k || n, but n doesn't divide (n/p^k)!, then we can use A1, since the maximum number of p-Sylows n/p^k. This takes out 24 (p=3), 36 (p=3), 48 (p=3), 80 (p=2), 96 (p=2), 100 (p=5).

Remaining: 12, 30, 56, 72, 90.

12. If there are 4 Sylow 3-groups, there's only room for 1 Sylow 2-group.
56. Same argument as 12. If there are 8 7-Sylows, there's only room for one 2-Sylow.

30. If there are 6 Sylow 5-groups, that leaves 6 elements over. So there aren't 10 3-Sylows, hence there's 1.

72. |Syl_3| = 1 or 4. If 4, we can use A1.

90. This is the only really tough one, and the proof I found uses several tricks we didn't discuss.

First, |Syl_5| = 1,6. If there are 6, let S be one of them. Then the normalizer N_G(S) is exactly the stabilizer of S under the conjugation action on Syl_5(G). Which is to say, we can use our |G| = |orbit| |stabilizer| formula to compute |N_G(S)| = 90/6 = 15.

We know 15-element groups very well: they're cyclic. Each one has exactly 8 elements of order 15 (and 4 of order 5, 2 of order 3, 1 of order 1). Now we use an extension of question #2, saying "no two cyclic groups can share generators". (Otherwise the generator would generate each one, and they'd be the same cyclic group.)

Also, each of these 15-element groups contains only one 5-group. So there really are 6 of them (not e.g. just one, containing all of the 5-Sylows). This accounts for a whopping 6*8 elements of order 15.

We now have 90-24-48 = 18 elements that aren't of order 5 or 15.

The number of 2-Sylows is 1,3,5,9,15, or 45. The first three cases are easy to handle with A1. So we can assume there are at least 9 elements of order 2, leaving only enough room for one 3-Sylow.

Ouch!