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Let $P$ be a normal sylow $p$-subgroup of a finite group $G$.

Since $P$ is normal it is the unique sylow $p$-subgroup.

I would like to say if $\phi$ is an automorphism then $\phi(P)$ is also a sylow $p$-subgroup. Then uniqueness would finish the proof. But is that true?

Does an automorphism of a group always send subgroups to subgroups of the same order?

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  • $\begingroup$ The answer to your question is yes. To see this, you know that identity is taken to identity by a homomorphism. And, a homomorphism is a structure preserving map. $\endgroup$ – user21436 Dec 22 '11 at 16:09
  • $\begingroup$ I changed the last statement of the question $\endgroup$ – user9352 Dec 22 '11 at 16:15
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    $\begingroup$ An automorphism (or/and an isomorphism) is a structure preserving bijection. Since the map is bijective, a set in the domain is taken to another of same order. $\endgroup$ – user21436 Dec 22 '11 at 16:19
  • $\begingroup$ @Kannappan: This is not a comment - it is an answer. $\endgroup$ – Martin Brandenburg Dec 22 '11 at 16:31
  • $\begingroup$ @Martin But, just in case OP wants to improve his question as was the case before. In any case, I have drafted an answer here. $\endgroup$ – user21436 Dec 22 '11 at 16:44
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Let $\phi:G \to G$ be the automorphism. We'll prove $\phi(P)$ is a $p$-Sylow Subgroup of $G$.

$\phi$ takes identity to itself:

For any $e_G \in G$, the identity in G, $\phi(e_G.e_G)=\phi(e_G)\phi(e_G)$ which proves the result.(from cancellation law in a group)

$\phi$ takes inverses to inverses:

Use the fact that $gg^{-1}=e_G$ for any $g \in G$. Do a computation similar to the one above.

$\phi$ takes subgroups to subgroups:

Let $H \leq G$. We intend to prove $\phi(H)$ is a subgroup. Use the 'lemma' we have proved before and verify the subgroup criterion (that $\phi(H)$ is closed under multiplication and inverses. )

Now, by one of my comments above, (in fact by just using the bijectivity of the map $\phi$, and by looking at its restriction to $H$), we'll prove that $|\phi(H)|=|H|$.

Note that the definition for two sets to be of same cardinality is that there exists a bijection between them.

So, You are through.

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  • $\begingroup$ These are just the broad steps and I assume OP will fill in the missing details himself. $\endgroup$ – user21436 Dec 22 '11 at 16:47
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Take any automorphism $f$, then $\#f(P)=\#P$ (because in particular $f$ is a bijection), and since also the image of subgroups (in this case $P$) is a subgroup, you conclude $f(P)$ is a p-sylow, and since you remarked:

Since $P$ is normal it is the unique Sylow $p$ subgroup.

That means $f(P)=P$,

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