Let $f: G_{1} \rightarrow G_{2}$ be a homomorphism. Show that if $H_{1} \leq G_{1}$, then $f\left(H_{1}\right) \leq G_{2}$ If $H_{1} \leqslant G_{1}$ then $f\left(H_{1}\right) \leqslant G_{2}$
I try:
Let $x,y \in H_1$, then $x,y \in G_1$ because $H_{1} \leqslant G_{1}$
$f(x*y)=f(x)\star f(y)$ because $f$ is an homomorphism and how $f(x),f(y)\in G_2 $
then $f(x)\star f(y) \in G_2$
and $f(x*x^{-1})=f(x)\star f(x^{-1})=f(x)\star f(x)^{-1}=e_2 \in G_2$
But I don't know how I can show that the inverse is in $G_2$.

From your notation, I'm assuming that $H_1$ is a subgroup of $G_1$, and $f$ is an homomorphism from $G_1$ and $G_2$.
Now, if $y=f(g)$ for some $g\in H_1$, then as $g^{-1}\in H_1$ and $y^{-1}=(f(g))^{-1}=f(g^{-1})$, one gets that $y^{-1}\in f(H_1)$.

Here we use the fact that if $f:G_1\rightarrow G_2$ is a homomorphism between groups, then $(f(g))^{-1}=f(g^{-1})$. This can be seen from
$$f(\mathbb{1}_{G_1})=\mathbb{1}_{G_2},$$
$$ \mathbb{1}_{G_2}=f(gg^{-1})=f(g)f(g^{-1}),$$
and
$$ \mathbb{1}_{G_2}=f(g^{-1} g)=f(g^{-1})f(g)$$
where $\mathbb{1}_{G}$ is the unit of group $G$.
