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.
 A: 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.
A: Looks correct. However, you can prove this more easily by noting that 
$$aha^{-1}=aga^{-1}\implies a^{-1}(aha^{-1})a=a^{-1}(aga^{-1})a\implies h=g$$
A: I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is Problem 4(b) on p.47.
I solved this problem as follows:

Let $\phi$ be a mapping such that $\phi:H\ni b\mapsto aba^{-1}\in aHa^{-1}$.
If $\phi(c)=\phi(d)$, then $c=a^{-1}\phi(c)a=a^{-1}\phi(d)a=d$.
So, $\phi$ is injective.
If $c\in aHa^{-1}$, then we can write $c=aha^{-1}=\phi(h)$ for some $h\in H$.
So, $\phi$ is surjective.
So, $o(H)=o(aHa^{-1})$.

