Subset of a finite group Let G be a finite group. Let $$H = \{b \in G.\ bab^{−1} \in \langle a \rangle \}.a \in G$$
Prove that if G is a finite group, then H is a subgroup of G.I think that a good approach is to prove that $$ba^i b^{-1} = (bab^{−1})^i \text{ for i an integer}$$
But I need a pointer on how to do that

Try this:
$H $ is not empty as $e_G \in H$ 
Let $b_1,b_2\in H$ then $b_1ab_1^{-1},b_2ab_2^{-1}\in \langle a \rangle $. So let $b_1ab_1^{-1}=a^m$, $b_2ab_2^{-1}=a^p$
Now $b_1b_2a(b_1b_2)^{-1}=b_1(b_2ab_2^{-1})b_1^{-1}=b_1a^m b_1^{-1}=(b_1ab_2^{-1})^m=a^{pm}\in \langle a\rangle$. Thus $b_1b_2\in H$.
Again $a=b^{-1}a^mb$. Since $G$ is a finite group and $b^{-1}ab \in G $ so $o(b^{-1}ab)=k$ (say).
So $(b^{-1}ab)^m=a\implies (b^{-1}ab)^{km+1}=a^{km+1}\implies 
b^{-1}ab=a^{km+1}\in \langle a\rangle $  
[Since $o(b^{-1}ab)=k\implies(b^{-1}ab)^{km+1}=b^{-1}ab$]
