Let $f:G \to H$ be surjective group homomorphism. Show that if $G=\langle a \rangle$, then $H=\langle f(a) \rangle$ 
Let $f:G \to H$ be surjective group homomorphism. Show that if $G=\langle a \rangle$, then $H=\langle f(a)\rangle$

Homorphism means $f(a*a)=f(a)*f(a)$
onto means $$\forall h \in H, \exists g \in G : f(g)=h $$
$H=\langle f(a)\rangle \Leftrightarrow H \subset \langle f(a)\rangle, 
\langle f(a)\rangle \subset H$

Assume $G=\langle a\rangle$
$\Rightarrow$ (if $h \in H$ is $h \in \langle f(a)\rangle$?)
$h \in H$ so $\exists g \in G$ such that $f(g)=h$ and $g\in \langle a \rangle$ so $\exists  k 
\in Z$ s.t $f(a^k)=f(a)^k=h$ so $h\in \langle f(a)\rangle$
$\Leftarrow$ Showing $\langle f(a)\rangle \subset H$)
let $x\in \langle f(a)\rangle $ so $\exists k ...$ $f(a)^k=x$ so $f(a^k)=x$ and $a^k \in G$  that is $f(a^k)\in H$

Critique?
 A: The idea of your proof is correct. Good job. :) As said in the comments though, the bit about proving that $\langle f(a) \rangle \subset H$ is unnecessary. I suppose you can write it down if you want, but it's kinda "obviously true": $H$ is the codomain of the map, so $f(a) \in H$, and since $H$ is a group any power of $f(a)$ must be in there too. 
In line with what users are saying in the comments, my critiques are about language/grammar. A major thing to remark on is that your proof looks like some math symbols with a few conjunction words thrown in. First and foremost, a proof should be a collection of sentences. Sentences that are grammatically correct, and that give your proof a sense of flow. Remember that your proof is meant to be read and understood by a person, and not just some grading robot looking for correctness ;) . I might write-up the same proof like this:

Assume $G=\langle a \rangle$, and consider some $h \in H$. Since $f$ is surjective, there must be some $g \in G$ such that $f(g)=h$. The group $G$ is generated by $a$, so there must be some integer power $k$ such that $g=a^k$. But $f$ is a homomorphism, so we have $$h = f(g) = f(a^k)=f(a)^k\,,$$ so $h\in \langle f(a)\rangle\,.$ That is to say, a general element $h \in H$ is a power of $f(a)$, so $H = \langle f(a) \rangle$.

Replacing certain mathematical symbols in lieu of their English language counterpart is often a good idea to help the readability of a proof. 
