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 $G_2$.
 A: 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 $G$.
A: I will use the one-step subgroup test.
Since $H_1\le G_1$, we have $e_1\in H_1$. Since $f$ is a homomorphism, $e_2=f(e_1)\in f( H_1)$, so $f(H_1)\neq\varnothing$.
By definition, $$f(H_1)=\{g\in G_2\mid \exists h\in H_1, g=f(h)\},$$ so we have $f(H_1)\subseteq G_2$.
Let $x,y\in f(H_1)$. Then there exist $a,b\in H_1$ such that $x=f(a)$ and $y=f(b)$. Now
$$\begin{align}
x\star y^{-1}&=f(a)\star (f(b))^{-1}\\
&=f(a)\star f(b^{-1})\\
&=f(a\ast b^{-1}),
\end{align}$$
but $a\ast b^{-1}\in H_1$ since $H_1\le G_1$. Hence $x\star y^{-1}\in f(H_1)$.
Hence $f(H_1)\le G_2$.
A: You want to show  that $f(H_1)\leqslant G_2$. You are confused about the inverse so, I am just solving that part. Note that under homomorphism, identity element $e_{G_1}\in G_1$ maps to identity in $G_2$. This is because $f(e_{G_1})=f(e_{G_1}.e_{G_1})=f(e_{G_1}).f(e_{G_1})\Rightarrow f(e_{G_1})=e_{G_2}$, by cancellation law. Hence, if we take $a\in G_1$, let $b\in G_1$ be the inverse of $a$, then $f(e_{G_1})=f(a.b)=f(a).f(b)=e_{G_2}$. Therefore, $f(b)$ is your inverse.
