# Prove that $g∶ H \to f(H),$ given by $g(x)=f(x)$ is an isomorphism, given $f$ is an isomorphism.

Prove that $$g∶ H \to f(H),$$ given by $$g(x) = f(x)$$ is an isomorphism, given $$f$$ is an isomorphism.

Let $$f∶ K \to G$$ be an isomorphism of groups. I also proved before this question that $$f(H) = \{y ∈ G ∶ y = f(x) \text{ for some } x \in H \}$$ is a subgroup of $$G$$.

So I have to show $$g$$ is a homomorphism, injective, and surjective. Let $$x,y \in H$$. Then $$g(xy)=f(xy)=f(x)f(y)=g(x)g(y).$$ (Since $$f$$ is a homomorphism). Let $$y=f(x) \in f(H),$$ and $$x \in H$$. Then $$g(x)=f(x)=y.$$ Hence it is surjective. I know that I could prove an injective homomorphism by using $$g(a)=e_{f(H)}$$ implies $$a=e_H$$. I have trouble with injectivity sometimes. Also is what I wrote for surjective ok?

Let $$h,k\in H$$. We have

\begin{align} g(hk)&=f(hk)\\ &=f(h)f(k)\\ &=g(h)g(k), \end{align}

so $$g$$ is an homomorpism.

Let $$y\in f(H)$$. Then there is some $$h\in H$$ with $$y=f(h)$$. But $$g(h)=f(h)$$. Hence $$g$$ is surjective.

Suppose $$g(h)=g(k)$$ for $$h, k\in H$$. Then

\begin{align} f(h)&=g(h)\\ &=g(k)\\ &=f(k). \end{align}

But $$f$$ is injective since it is an isomorphism. Hence $$h=k$$. Hence $$g$$ is injective.

Hence $$g$$ is an isomorphism.

• OOOOOOOooooo...yes I see haha. Thank you!! Apr 19, 2020 at 0:12
• You're welcome, @PhysicsBish :) Apr 19, 2020 at 0:14