Proof question: Show there is no homomorphism from $\Bbb Z_{16} \oplus \Bbb Z_{2}$ onto $\Bbb Z_4 \oplus \Bbb Z_4$

Question

Prove that there is no surjective homomorphism from $\Bbb Z_{16} \oplus \Bbb Z_{2}$ onto $\Bbb Z_4 \oplus \Bbb Z_4$.

The book I'm reading starts off the proof with:

Suppose $\phi$ is such a homomorphism between the two groups, then $f: \Bbb Z_{16} \oplus \Bbb Z_{2} / \ker \phi \rightarrow \phi(\Bbb Z_{16} \oplus \Bbb Z_{2})$ is an isomorphism. Therefore $\ker \phi = \langle (8,1) \rangle, \langle (0,1 \rangle),$ or $\langle (8,0)\rangle$.

How does the isomorphism imply that this is what the kernel equals?

Answer 1

Note that two isomorphic groups have the same number of elements. So, $|\Bbb Z_{16} \oplus \Bbb Z_2/\ker \phi| = 16$. So, $|\Bbb Z_{16}\oplus \Bbb Z_2|/|\ker \phi| = 16$.

So, $|\ker \phi| = 2$.

Answer 2

The order of the domain is twice that of the range, so if the homomorphism is surjective, then the kernel has order $2$. That last sentence just listed all of the subgroups of the domain of order $2$.

Answer 3

The cardinality of $Z_{16}\times Z_2$ is 32 and the cardinality of $Z_4\times Z_4$ is 16. Thus the cardinality of the kernel of $\phi$ is 2 since it is surjective. The only subgroups of order 2 of $Z_{16}\times Z_2$ are generated by $(8,1), (0,1), (8,0)$.