# A Sylow Subgroup Question

1. Nov 2, 2008

### e(ho0n3

The problem statement, all variables and given/known data
If G is a group of order 231, prove that the 11-Sylow subgroup is in the center of G.

The attempt at a solution
The number of 11-Sylow subgroups is 1 + 11k and this number must be either 1, 3, 7, 21, 33, 77 or 231. Upon inspection, the only possibility is 1.

Let H be the 11-Sylow subgroup of G. Note that 3 * 7 * 11 = 231, so o(H) = 11 and H is cyclic. Furthermore, this is the only subgroup of order 11, so it must be normal. With these facts in mind, let a be a nonidentity element of H. For any g in G, we must have g-1ag = aj for some j in {1, 2, ..., 11}. Thus ag = gaj.

I don't know what to do next. Any tips?

2. Nov 3, 2008

### morphism

What can you say about the order of the automorphism g -> g-1ag? (Hint: it's enough to consider those g's coming out of the 3- and 7-Sylows.)

3. Nov 3, 2008

### e(ho0n3

If the order of g is 3 or 7, then the order of the automorphism must be 21. But now what?

4. Nov 3, 2008

### morphism

Why do you say that?

By the way, there was a typo and an unfortunate choice of letters in my previous post. But I hope the idea got through though - we want to let G act on H by conjugation. That is, consider the automorphism f_g of H defined by f_g(x) = g-1xg. What is its order? (Hint: the map g->f_g is a homomorphism.)

5. Nov 4, 2008

### e(ho0n3

I said 21 because because it is the least common multiple of 3 and 7. Thus, applying f_g 21 times to x will surely result in just x.

6. Nov 4, 2008

### morphism

Sure, but does that mean the order of f_g is 21?

Here's another hint. If the order of g is 3, then the map g->f_g is an embedding from Z_3 into Aut(H)=~Z_10.

7. Nov 4, 2008

### e(ho0n3

No. Let n be the order of fg. Then g-nagn = a or agn = gna. If g in H, then n must be 1. If g is in a 3-Sylow subgroup, then n must be 3 unless g commutes with elements from H. Is this possible?

Wait. Where did you get that Aut(H) =~ Z10? I thought Aut(H) =~ U11.

8. Nov 5, 2008

### morphism

If U_11 is the group of invertibles of Z_11, then it's isomorphic to Z_10 (it's cyclic and has order phi(11)=10).

9. Nov 5, 2008

### e(ho0n3

Oh yeah. Still, I don't see where you're going with this.

10. Nov 5, 2008

### morphism

Well, a subgroup is central iff conjugating it is a trivial action. Since H itself is abelian, we only need to worry about conjugation with elements outside of H. By Lagrange, the non-identity elements in G\H will lie in either the 3-Sylow or 7-Sylow.

Let's consider the 3-Sylow, call it P. The map g -> f_g is a homomorphism from P into Aut(H)=Z_11. An order argument will yield that f_g=1. Do the same for the 7-Sylow.

11. Nov 6, 2008

### e(ho0n3

But the elements of G/H are cosets. I don't get.

You meant to write Z_10. I understand know. I did not suspect the order of f_g to be 1. I'll have to think about that.

12. Nov 6, 2008

### e(ho0n3

I just realized you meant G - H when you wrote G\H. Sorry about that. Still, I don't see how Lagrange implies that though.

13. Nov 6, 2008

### e(ho0n3

Since Aut(H) is isomorphic to Z10, it has 10 elements, whence the order of fg must be either 1, 2, 5 or 10. And since we know that (fg)3 is the identity map (assuming g is an element of a 3-Sylow subgroup), we can rule out 2, 5, and 10, so its order must be 1.

14. Nov 6, 2008

### morphism

Yup.

You can disregard this bit:
It's obviously nonsense. I don't know why I wrote it.

Anyway, to finish the argument, notice that f_g=1 for every g in P means that P is a subset of the centralizer of H. So...

15. Nov 7, 2008

### e(ho0n3

I understand the rest. Thank you. However, we still have to show that every non-identity element in G - H has order 3 or order 7.

16. Nov 7, 2008

### morphism

No we don't. It's actually not true (think of an abelian counterexample). Initially I 'broke up' the group into elements of order 3, 7 and 11 to guide my intuition, but then I took it too far.

The point is, we can embed subgroups of order 3, 7 and 11 into the centralizer of H. So the centralizer must be the entire group. That's all there is to it.

17. Nov 7, 2008

### e(ho0n3

I'm not sure I understand what you mean by "embed" or how that proves that C(H) = G. We have shown that if g has order 3 or 7, then it belongs to C(H). Does the same argument we used to show that the order of fg is 1 work for those g that have order 21, 33, or 77? This seems to be the case.

18. Nov 9, 2008

### e(ho0n3

I've been thinking, if f_g is an automorphism of H, then g must be in H and we can't really use it if g is not in H, which destroys the whole argument we've developed.

19. Nov 9, 2008

### e(ho0n3

Oh wait. I forgot that H is normal so g-1xg is in H for all x in H and g in G.

20. Nov 9, 2008

### e(ho0n3

OK. I've written down a solution which I'm content with. Thanks for the help morphism.