The Question "Prove that there is no homomorphism from Z16 ⊕ Z2 onto Z4 ⊕ Z4" In texbook, Is my proof correct? Assume，φ: Z16 ⊕ Z2 → Z4 ⊕ Z4, φ is HOMO. onto.
|Z16 ⊕ Z2| = |Z16||Z2| = 32 > 16 = |Z4||Z4| = |Z4 ⊕ Z4|
⇒
∃(a,b),(c,d) ∈ Z16 ⊕ Z2, (a,b) ≠ (c,d), φ((a,b)) = φ((c,d))
⇒
φ((a,b) + (c,d)) = φ((a + c, b + d)) =
2φ((a,b)) = φ(2(a,b)) = φ((2a,2b)) =
2φ((c,d)) = φ(2(c,d)) = φ((2c,2d))
⇒
a + c = 2a, b + d = 2b,
a + c = 2c, b + d = 2d,
⇒
a = c, b = d
⇒
(a,b) = (c,d)（contradiction）
 A: I am not pure algebraist mathematician, but I thought an ideal solution would be in the following way:
If $φ$ is onto, ${\rm Ker}(φ)<\Bbb{Z}_{16}\oplus\Bbb Z_2$ is a normal subgroup of order $\frac{16\times 2}{4\times 4}=2$. So, we have $3$ cases:
i) ${\rm Ker}(φ)=\Bbb Z_2\oplus 0$ in which case ${\rm Im}(φ)\cong\frac{\Bbb{Z}_{16}\oplus\Bbb Z_2}{\Bbb Z_2\oplus 0}\cong\Bbb{Z}_{8}\oplus\Bbb Z_2$,
ii) ${\rm Ker}(φ)=0\oplus\Bbb Z_2$ in which case ${\rm Im}(φ)\cong\frac{\Bbb{Z}_{16}\oplus\Bbb Z_2}{0 \oplus \Bbb Z_2}\cong\Bbb{Z}_{16}$,
iii) ${\rm Ker}(φ)=D$, the diagonal $\Bbb Z_2$ in which case ${\rm Im}(φ)\cong\frac{\Bbb{Z}_{16}\oplus\Bbb Z_2}{D}\cong\Bbb{Z}_{16}$ again.
Contradiction.
A: I think OP is seeking for an elemantry proof which uses only definition of HOMO. and some modular arithmetic avoiding some isomorphism theorems, kernels or quotients, etc.
As pointed out J. Lahtonen, by definition of group homomorphism, we notice that $$\phi(4a,4b)=4\phi(a,b)=(0,0).$$ Hence, we reach the important conclusion that $\phi(0,0)=\phi(4,0)=\phi(8,0)=\phi(12,0)=(0,0).$ Then, we conclude that the image is determined by the elements $$\{(0,0),(1,0),(2,0),(3,0),(0,1),(1,1),(2,1),(3,1)\}\subset\Bbb Z_{16}\oplus\Bbb Z_2$$ and the image can contain at most $8$ elements. Contradiction.
