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

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.

• $f(x)f(t^{-1}) = f(xt^{-1})$ Apr 17, 2020 at 22:11
• You are almost there: $f(x)f(t)^{-1}=f(x)f(t^{-1})=f(xt^{-1})$. Now, as $x, t\in H$ and $H$ is a subgroup, then $xt^{-1}\in H$ and so $f(xt^{-1})\in f(H)$... Apr 17, 2020 at 22:12
• Ahhhh yes using the homomorphism properties again. Thanks @Riccardo Allegrone Apr 17, 2020 at 22:23
• and @Stinking Bishop, thank you! Apr 17, 2020 at 22:23
• Note that $f(e_K)\in H$ since $f$ is an isomorphism (and hence an injective homomorphism), so $f(H)\neq \varnothing$. Apr 17, 2020 at 22:54

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 $$f(xt^{−1})\in f(H)$$.
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.