No homomorphism from $Z_{16}\oplus Z_{2}$ onto $Z_{4}\oplus Z_{4}$.

For some reasons, homomorphism is a very hard area for me to make improvements. I've been hitting the brick wall for almost 2 hours.

Prove that no homomorphism exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$.

Assume a homomorphism $\phi$ exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$. Then $\phi$ is an isomorphism.

Note that the $\left | \ker \phi \right |=2$ The $\ker \phi$ is also a normal subgroup for $Z_{16}\oplus Z_{2}$. We want possible normal subgroup of order 2. I.e., 2 elements in each normal subgroups.

By Lagrange's theorem, the order of each element in a group divides the order of a group so the possible order of the elements are 1 or 2. If the elements has order 1, then the Ker \phi cannot have 2 elements.

Hence, $\ker\phi$ = $\left \{ (0,0),(8,0) \right \}$ ,$\left \{ (0,0),(0,8) \right \}$,$\left \{ (0,0),(8,1) \right \}$, possibly.

By the First isomorphism theorem:

$\phi: Z_{16}\oplus Z_2 \rightarrow Z_2\oplus Z_2$

$\Psi: G/\ker \phi \rightarrow \phi\left ( Z_16\oplus Z_2 \right )=Z_2\oplus Z_2$

$g\ker \phi \mapsto \phi(g)=\Psi(g\ker \phi)$

Any help is appreciated.

Edit: I know that $(2,0)$ has order $8$ in $Z_{16}\oplus Z_{2}$ but any elements in $Z_{4}\oplus Z_{4}$ does not have order $8$ which would have solved the question at the outset. But I would like a different route.

Solution 1:

Let $f:Z_{16}\times Z_2\to Z_4\oplus Z_4$ be a map. The subgroup $f(Z_{16})$ is cyclic, so it has at most $4$ elements (as $4$ is the maximum order of an element in the codomain of $f$) Now the image of $f$ is equal to the subgroup $f(Z_{16})+f(Z_2)$, which has at most $8$ elements. The map $f$ is therefore not surjective.