Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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. – user21436 Dec 22 '11 at 16:09
I changed the last statement of the question – user9352 Dec 22 '11 at 16:15
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. – user21436 Dec 22 '11 at 16:19
@Kannappan: This is not a comment - it is an answer. – Martin Brandenburg Dec 22 '11 at 16:31
@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. – user21436 Dec 22 '11 at 16:44
up vote 2 down vote accepted

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.

share|cite|improve this answer
These are just the broad steps and I assume OP will fill in the missing details himself. – user21436 Dec 22 '11 at 16:47

Your Answer


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.