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.

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.