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?

• Use $\langle\rangle$ for $\langle\rangle$. – Shaun Sep 28 '17 at 15:05
• Your proof seems fine to me. – Shaun Sep 28 '17 at 15:08
• The second step doesn't require any proof : $\langle f(a)\rangle \subset H$ holds by definition. You might also be interested in this question : math.stackexchange.com/questions/1123005/… – Arnaud D. Sep 28 '17 at 15:10
• The only critical remark I can make is that "cruitique" is normally spelled c-r-i-t-i-q-u-e.! – Robert Lewis Sep 28 '17 at 15:10
• Perhaps you could reconsider your username? – Alex Clark Sep 28 '17 at 15:21

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.
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 $H = \langle f(a) \rangle$.