Prove that $f(H)=\{y\in G∶y=f(x)\text{ for some }x\in H\}\le G.$

Question

Let $f∶ K \rightarrow G$ be an isomorphism of groups and Suppose $H$ is a subgroup of $K.$

So, I will do the subgroup test. Show its nonempty and that $ab^{-1} \in f(H).$ Maybe let $a \in f(H),$ then $a=f(x)$ for some $x\in H$ and $b \in f(H),$ then $b=f(t)$ for some $t \in H.$ Is that right? If so then : $ab^{-1}= f(x)f(t)^{-1}$, and since $f$ is a homomorphism, $f(x)f(t^{-1})$. But I can't figure out the next step if this is what I should be doing. I have thoughts that since they are groups that the inverses are elements too, so $f(t^{-1})$ is also in $f(H)$? Hm...not too sure.

Answer 1

You are almost there: $f(x)f(t)^{−1}=f(x)f(t^{−1})=f(xt^{−1})$. Now, as $x,t∈H$ and $H$ is a subgroup, then $xt^{−1}\in H$ and so $f(xt^{−1})\in f(H)$.

Answer 2

Note that all you need is $f$ to be a group homomorphism, as the injectivity and surjectivity of $f$ (as a map with codomain $G$!) don't play any role in proving the claim. In fact, $e_K\in H$ and $f(e_K)=e_G$, so $f(H)\ne\emptyset$. The rest is okay as you (almost) did and the other answer has completed.