Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is one part of a homework question. If we show this fact, then the rest of the problem is solved.

Let $G$ be a finite group and let $H$ be the subgroup generated by all Sylow p-subgroups. We want to show $H\lhd G$. Here is my reasoning so far:

At first, I thought that $H$ contained all of and only the elements of $G$ of order $p^k$ for some $k>0$. If this were the case, then $ghg^{-1}$ would have the same order as $h$, that is, $p^k$ for some $k$. That would mean that $ghg^{-1}\in H$ so we would be done if we could show that $H$ is not a p-group. For, if it were a p-group, it would contain a Sylow p-subgroup of $G$, a contradiction.

Also, my first statement about $H$ containing all of and only the elements of $G$ is suspicious. Any help would be appreciated. Thank you.

share|improve this question
add comment

3 Answers

up vote 3 down vote accepted

Any automorphism of $G$ permutes the Sylow $p$-subgroups of $G$. Therefore, $H$ must be fixed by any automorphism. Thus, $H$ is a characteristic subgroup of $G$. In particular, it is normal in $G$.

share|improve this answer
add comment

Unfortunately, the statement you want to use to prove the result isn't the case. For example, if $p = 2$ and $G = S_3$, the subgroup generated by all the Sylow 2-subgroups is the subgroup generated by all the transpositions, which is $G$ itself, so in particular has elements of order 3 in it.

A hint about how to show the result: all Sylow $p$-subgroups are conjugate. So if you have $$ w = g_1g_2\ldots g_n $$ in the subgroup generated by the Sylow $p$-subgroups, where each $g_n$ is itself actually in a Sylow $p$-subgroup, and you take $kwk^{-1}$, you should be able to rewrite this as a product of $h_i$s where each $h_i$ is in some Sylow $p$-subgroup (maybe not the same one as $g_i$). You will need to use a mild "trick" to do this.

share|improve this answer
    
Thanks. That will certainly do. –  Anri Rembeci Oct 26 '12 at 2:58
add comment

Another approach: let $\,X_p:=\{P\leq G\;\;;\;\;P\,\,\text{is a Sylow}\,\,p-\text{subgroup}\}\,$ .

Now, by Sylow theorems we know that

$$\forall \,x\in G\,\,\forall\,P\in X_p\;\;,\;P^x:=x^{-1}Px\in X_p\Longrightarrow \langle\,X_p\,\rangle^x=\langle\,X_p\,\rangle\Longrightarrow \,\langle\,X_p\,\rangle\triangleleft G$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.