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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.

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up vote 4 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 $G$.

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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.

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Thanks. That will certainly do. – Anri Rembeci Oct 26 '12 at 2:58

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

Now, by Sylow theorems we know that

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

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