# Conjugate subgroups.

Printable View

• Nov 26th 2009, 11:43 AM
platinumpimp68plus1
Conjugate subgroups.
Let H be a subgroup of G, with index p (prime). What are the possible numbers of conjugate subgroups of H?

I'm pretty lost. I think it has something to do with Sylow's Theorems, and the fact that if p=|G|/|H|, so then p divides|G| and there is a p-Sylow subgroup, and Sylow-p's are conjugate to each other? But then I get really confused and forget what I'm trying to do, and the question didn't say G is finite so I don't even know if that method applies. Any help?
• Nov 26th 2009, 03:01 PM
tonio
Quote:

Originally Posted by platinumpimp68plus1
Let H be a subgroup of G, with index p (prime). What are the possible numbers of conjugate subgroups of H?

I'm pretty lost. I think it has something to do with Sylow's Theorems, and the fact that if p=|G|/|H|, so then p divides|G| and there is a p-Sylow subgroup, and Sylow-p's are conjugate to each other? But then I get really confused and forget what I'm trying to do, and the question didn't say G is finite so I don't even know if that method applies. Any help?

I don't think there's one unique answer to this question, which is most probably directed to finite groups (read at the beginning of the chapter/section whether all the groups in that chapter/section are assumed to be finite).
Now, if p was the MINIMAL prime that divides the order of p then we have a very simple and nice answer...

Tonio
• Nov 26th 2009, 03:05 PM
platinumpimp68plus1
Quote:

Originally Posted by tonio
I don't think there's one unique answer to this question, which is most probably directed to finite groups (read at the beginning of the chapter/section whether all the groups in that chapter/section are assumed to be finite).
Now, if p was the MINIMAL prime that divides the order of p then we have a very simple and nice answer...

Tonio

Hmmm, okay. It didn't say p was the smallest prime but I'm think it is supposed to be assumed, just based on other problems in the chapter. I don't know how to use that though... :(
• Nov 26th 2009, 03:09 PM
tonio
Quote:

Originally Posted by platinumpimp68plus1
Hmmm, okay. It didn't say p was the smallest prime but I'm think it is supposed to be assumed, just based on other problems in the chapter. I don't know how to use that though... :(

Then H is normal in G...(Wink) . The prove passes through the action of G on the set of left cosets of G in H, and then using the associated homomorphism's kernel and etc.

Tonio
• Nov 26th 2009, 03:27 PM
platinumpimp68plus1
Quote:

Originally Posted by tonio
Then H is normal in G...(Wink) . The prove passes through the action of G on the set of left cosets of G in H, and then using the associated homomorphism's kernel and etc.

Tonio

Thanks... I'm still really confused though. I'm pretty lost in general when it comes to G-sets and conjugate groups. I dunno how to start/how this tells me how many conjugate subgroups of H I have. Could you explain in some more detail?
• Nov 26th 2009, 04:11 PM
tonio
Quote:

Originally Posted by platinumpimp68plus1
Thanks... I'm still really confused though. I'm pretty lost in general when it comes to G-sets and conjugate groups. I dunno how to start/how this tells me how many conjugate subgroups of H I have. Could you explain in some more detail?

If p is the minimal prime that divides the order of G, or if simple it is given H is a normal sbgp. of G, then all the conjugates of H are...H itself, of course!

Tonio
• Nov 26th 2009, 04:35 PM
aliceinwonderland
Quote:

Originally Posted by platinumpimp68plus1
Let H be a subgroup of G, with index p (prime). What are the possible numbers of conjugate subgroups of H?

I'm pretty lost. I think it has something to do with Sylow's Theorems, and the fact that if p=|G|/|H|, so then p divides|G| and there is a p-Sylow subgroup, and Sylow-p's are conjugate to each other? But then I get really confused and forget what I'm trying to do, and the question didn't say G is finite so I don't even know if that method applies. Any help?

If $\displaystyle H \leq G$, then the number of conjugates of H in G is equal to the index of its normalizer, i.e., $\displaystyle [G:N_G(H)]$.

Since H is a maximal subgroup of G, $\displaystyle N_G(H)$ is either H or G.
If $\displaystyle N_G(H)=H$, then the number of conjugates of H in G is p. If $\displaystyle N_G(H)=G$, then the number of conjugates of H in G is 1.

If you need to prove that the number of conjugates of H in G is equal to the index of its normalizer, you need to define a function $\displaystyle f:[H] \rightarrow G/N$ given by $\displaystyle f(aHa^{-1})=aN$, where [H] denotes the family of conjugates of H and N denotes $\displaystyle N_G(H)$. I'll leave it to you to verify that f is a bijection.