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?


1 Answer 1


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.

  • 1
    $\begingroup$ OOOOOOOooooo...yes I see haha. Thank you!! $\endgroup$ Apr 19, 2020 at 0:12
  • $\begingroup$ You're welcome, @PhysicsBish :) $\endgroup$
    – Shaun
    Apr 19, 2020 at 0:14

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