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.