Let $G$ be a group. Suppose that the map $\Phi : G \to G$ given by $\Phi(a) = a^3$ is a group homomorphism, prove that
For every $a, b \in G$, $baba = a^2b^2$.
The subset $H = \{a^2| a \in G\}$ is a subgroup of $G$.
I have already proven the first one but I do not know how to prove the second one. Can anyone help to prove the second one? I know that I have to show that it contains a identity element, inverse and is closed under group operation. Thank you guys in advance.
The inverse of $a^2$ is $(a^{-1})^2$, and this is in $H$ by definition. The identity is also in $H$ since the identity times itself gives the identity. Closure under multiplication follows from $(1)$ since $a^2b^2=baba=(ba)^2$ and $ba$ is in $G$, thus $a^2b^2$ is in $H$.