Show that $\Bbb Z_{16}$ is not a homomorphic image of $\Bbb Z_4×\Bbb Z_4$.
My solution goes like this:
We consider $f$ as an epimorphism. Now, $f:\Bbb Z_4×\Bbb Z_4\longrightarrow \Bbb Z_{16}$. Hence, $\Bbb Z_4×\Bbb Z_4/Kerf\cong \Bbb Z_{16}$. So, $\frac{|\Bbb Z_4×\Bbb Z_4|}{|Kerf|}=|\Bbb{Z}_{16}|$. This means $Kerf=1$, and so, $f$ is an isomorphism, so $\Bbb Z_4×\Bbb Z_4\cong \Bbb Z_{16}$. Now, $\Bbb Z_{16}$ is cyclic, while $\Bbb Z_4×\Bbb Z_4$, is not. So, they are not isomorphic, hence, contradiction.
Is this a correct argument? Is the proof correct?
Alternatively (without the homomorphism theorem), let $f\colon G\to H$ be a homomorpism of finite groups. Then, for every $x\in G$: \begin{alignat}{1} e_H=f(e_G)=f(x^{o(x)}) &= f(x)^{o(x)} \end{alignat} and hence: $$o(f(x))\mid o(x)\tag1$$
In your setting, if $f$ is surjective, then an element of order $16$ of $\Bbb Z_{16}$ is the image of some $x\in\Bbb Z_4\times\Bbb Z_4$, whose order is less than $16$ (because the elements of this group have maximal order $4$). But this contradicts $(1)$. Therefore, there isn't any such a surjective homomorphism.