Prove function is a homomorphism

Question

I'm having trouble trying to prove the following:

Let $G$ and $H$ be two groups and let $f:G\to H$, $g:H\to G$. Show that if $g$ is injective and $g$ and $g\circ f$ are homomorphisms, then $f$ too, is a homomorphism.

This is what I've done so far:

Suppose $g:H\to G$ is injective. Then,

$\forall h_1,h_2 \in H$, from $g(h_1)=g(h_2)\implies h_1=h_2$

Now, $\forall g_1,g_2 \in G$ we have:

$\begin{aligned} f(g_1g_2)&=f(g(h_1)g(h_2))\\ &=f(g(h_1h_2)) \\&=f\circ g (h_1h_2) \end{aligned}$

And this is where I'm stuck.

Answer 1

First step: show $f(e_G) = e_H$: Let $h ∈ H$. Then $$ \begin{align} g( f(e_g) h) &= g(f(e_g)) g(h) & g \text{ Homom.} \\ &= e_G g(h) & g ∘f \text{ Homom.} \\ &= g(h) & e_G \\ ⇒ f(e_g)h &= h & g \text{ injective} \end{align} $$ Do the same for $h f(e_G)$ and you have that $f(e_G)$ is the neutral element of $H$: $e_h$.

Similarly we get the other homomorphism property of $f$. Let $g_1, g_2 ∈ G$. Then $$ \begin{align} g ( f (g_1 g_2) ) &= (g∘f)(g_1 g_2) & \text{Def. } ∘ \\ &= (g ∘ f)(g_1) (g∘f)(g_2) & g∘f \text{ Homom.} \\ &= g(f(g_1)) g(f(g_2)) \\ &= g(f(g_1)f(g_2)) & g \text{ Homom.} \\ ⇒ f(g_1g_2) = f(g_1)f(g_2) & g \text{ injective} \end{align} $$ Done.

Actually, if you continue your proof by using that $f∘g$ is a homomorphism and replacing $g(h_i)$ back by $g_i$ ($i = 1,2$) you also get the result that $f$ is a homomorphism. But you used that $g$ is surjective instead of injective. So in your theorem you have to assume either $g$ injective or surjective.

Answer 2

Let $x_1,x_2\in H$. because $f\circ g$ is homomorphism than $g(f(x_1\cdot x_2))=g(f(x_1))\cdot g(f(x_2))$ and because $g$ is homomorphism it follows that $g(f(x_1))\cdot g(f(x_2))=g(f(x_1)\cdot g(f(x_2))$ thus $g(f(x_1\cdot x_2))=g(f(x_1)\cdot f(x_2))$. because $g$ is injective than we can conclude that $f(x_1\cdot x_2)=f(x_1)\cdot f(x_2)$.