I have the following problem:
Let $\theta : G \rightarrow H$ be an onto group homomorphism and let $g$ $\in$ $H$. Define $A$ = {$x$ $\in$ $G$ : $\theta(x) = g$}. Prove that $A$ = $a(ker\theta)$ for every $a \in A$.
So far I have attempted the proof by doing two cases, one showing that A is a subset of $a(ker\theta)$ and the other showing that $a(ker\theta)$ is a subset of A. I've completed the first case, but I'm stuck on the second one (showing that $a(ker\theta)$ $\subseteq$ A). Any suggestions are appreciated!
Suppose $x\in \operatorname{ker} (\theta)$, and $a\in A$, then $\theta(ax)=\theta(a)\theta(x)=ge=g$, which implies $ax\in A$. So $a\operatorname{ker}(\theta)\subseteq A$.
Also if $b\in A$, then $a^{-1}b\in\operatorname{ker}(\theta)$ [because $\theta(a^{-1}b)=\theta(a^{-1})\theta(b)=(\theta(a))^{-1}g=e$], or $b\in a \operatorname{ker}(\theta)$. So $A\subseteq a \operatorname{ker}(\theta)$