Homomorphism of groups, subgroups

Question

If $f: G_1\to G_2$ is homomorphism of groups $G_1$ and $G_2$ and if $|G_2|=25$ and $A$ is subgroup of $G_1$ such that $A\neq \{e\}$, and $f(A)\neq G_2$, prove that $f(A)$ is a subgroup of $G_2$, and find the order of $G_2$.

Answer 1

Since $f$ is a homomorphism, $f(e_{G_1})=e_{G_2}$. So, $e_{G_2}\in f(A)$ and also, $f(A)$ is not empty.

Also, you can show that for a homomorphism, $f(g^{-1})=f(g)^{-1}$. This means that $f(gh^{-1})=f(g)f(h^{-1})=f(g)f(h)^{-1}$. So, for all $f(g),f(h) \in f(A)$, we have that $f(g)f(h)^{-1} \in f(A)$. This, together with the fact that $f(A)$ contains the identity, proves that $f(A)$ is a subgroup of $G_2$.

Now that we know it's a subgroup, by Lagrange's theorem, we know that $o(f(A)) \mid o(G_2)=25$. Since $f(A) \neq \{e\}$, we should have that $o(f(A))=5 \text{ or } 25$. Since $f(A) \neq G_2$, it can't be $25$ and therefore $o(f(A))=5$.