Group Homomorphism: Is it required to have the inverse preservation to prove Identity Mappings?

Question

Consider two Groups $\left(G,\cdot\right)$ and $\left(H,\times\right)$

And let $f$ be a function $f:G\rightarrow H$

Then $f$ is a Group Homomorphism if $\forall g_{1}, g_{2} \in G$ we have

$f\left(g_{1} \cdot g_{2}\right) = f\left(g_{1}\right) \times f\left(g_{2}\right)$

and that the inverse of $G$ is preserved, i.e.

$f\left(g_{1}^{-1}\right) = f\left(g_{1}\right)^{-1}$

In the textbook I'm reading this is essential so that the identity for $G$ here $1_{G}$ maps onto the identity of $H$ here $1_{H}$ .

This is illustrated simply with

$f\left(g_{1} \cdot g_{1}^{-1}\right) = f\left(1_{G}\right) = f\left(g_{1}\right) \times f\left(g_{1}^{-1}\right) = f\left(g_{1}\right) \times f\left(g_{1}\right)^{-1} = 1_{H}$

And so,

$f\left(1_{G}\right) = 1_{H}$

I however went about it differently (I'm sure I'm missing something) though and I was wondering if people could have a look over and see if what I've done is valid and if not where I went wrong.

So my method was

$f\left(1_{G}\cdot 1_{G}\right) = f\left(1_{G}\right) = f\left(1_{G}\right) \times f\left(1_{G}\right)$

And so

$f\left(1_{G}\right)^{-1} \times f\left(1_{G}\right) = f\left(1_{G}\right)^{-1} \times f\left(1_{G}\right) \times f\left(1_{G}\right)$

$1_{H} = 1_{H} \times f\left(1_{G}\right) = f\left(1_{G}\right)$

And so

$1_{H} = f\left(1_{G}\right)$

Is this valid? as in can we show that the identity of $G$ maps to the identity of $H$ without having to impose the inverse requirement as before?

Thanks

Answer 1

Your method looks correct. It's a fact that a group homomorphism preserves the identity element.

Answer 2

Yes. You are right and you proved correctly that if $f\colon G\longrightarrow H$ is such that $(\forall g_1,g_2\in G):f(g_1g_2)=f(g_1)f(g_2)$, then $f(1_G)=1_H$.