# order of $aHa^{-1}$

Can anyone please tell me if this proof is correct?

Question: If $H$ be a finite subgroup of $G$ and $a\in G$, let $aHa^{-1}=\{aha^{-1}|h\in H\}$. What is the order(or cardinality) of $aHa^{-1}?$.

Here is my attempt:

Lemma: $aHa^{-1}$ is a subgroup of $G$.

Proof: Every element of $aHa^{-1}$ is in $G$, so $aHa^{-1}\subset G$.

For $h_1,h_2\in H$, $(ah_1a^{-1})(ah_2a^{-1})=ah_1h_2a^{-1}=aha^{-1}$ as $h_1.h_2=h\in H$ for some $h\in H$ because $H$ is a subgroup of G. $(aha^{-1})(ah^{-1}a^{-1})=e$ where $E$ is the identity element of $H,G$ and $aHa^{-1}$.Thus $aHa^{-1}$ is a subgroup of $G$. $\blacksquare$

Let the order of $aHa^{-1}$ be denoted by $o(aHa^{-1})$. Clearly, $o(aHa^{-1})=o(Ha^{-1})$.We know that tthe cardinality of the right cosets of $H$ is $o(H)$.So, $o(aHa^{-1})=o(H)$

Please comment if my proof is correct.

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Where do you ever use the lemma that $aHa^{-1}$ is a subgroup? Why not just start at the last paragraph "Let the order of..."? – Erick Wong Sep 20 '12 at 5:41
Perhaps there was no need to do that but I was thinking of order as in order of a group,hence the trouble. – user41685 Sep 20 '12 at 5:55
True that seems reasonable but I noticed you refer to $o(Ha^{-1})$ even $Ha^{-1}$ is not a group. If it's obvious to you (as in you can see how to prove it) that $|aHa^{-1}| = |Ha^{-1}|$ then it's equally obvious that $|Ha^{-1}|=|H|$. – Erick Wong Sep 20 '12 at 13:47
You are right.As I said, that bit was unnecessary. – user41685 Sep 20 '12 at 13:54
@ErickWong In the last part why is $o(aHa^{-1}) = o(Ha^{-1})$ – Taylor Ted Mar 25 '15 at 4:37

Try to avoid using the word "clearly", even though your statement is correct. Why is it true that $|aHa^{-1}|=|Ha^{-1}|$? A better way to show this is to build a map $\phi \colon H \to aHa^{-1}$ given by $\phi(h)= aha^{-1}$. This is surjective, since any $aha^{-1} \in aHa^{-1}$ is mapped to by $\phi(h)$. It is injective, since $ah_1a^{-1}=ah_2a^{-1}$ implies $h_1=h_2$ by cancellation. Thus $\phi$ is a bijection and the two subgroups have the same order.
Looks correct. However, you can prove this more easily by noting that $$aha^{-1}=aga^{-1}\implies a^{-1}(aha^{-1})a=a^{-1}(aga^{-1})a\implies h=g$$