# Homomorphism from $\Bbb Z_8 \oplus \Bbb Z_2$ onto $\Bbb Z_4 \oplus \Bbb Z_4$?

Prove that there is no Homomorphism from $\Bbb Z_8 \oplus \Bbb Z_2$ onto $\Bbb Z_4 \oplus \Bbb Z_4$?

Suppose that $\phi$ is an onto homomorphism between the two sets.

Then $\phi(\Bbb Z_8 \oplus \Bbb Z_2) = \Bbb Z_4 \oplus \Bbb Z_4$ because it's onto and $|\phi(\Bbb Z_8 \oplus \Bbb Z_2)| = |\Bbb Z_4 \oplus \Bbb Z_4| = 16$.

Then $(\Bbb Z_8 \oplus \Bbb Z_2) / \ker\phi \approx \Bbb Z_4 \oplus \Bbb Z_4$.

Then $|(\Bbb Z_8 \oplus \Bbb Z_2) / \ker\phi| \approx |\Bbb Z_4 \oplus \Bbb Z_4| \Rightarrow |\ker\phi| = 1$.

and this implies that the homomorphism is injective and onto, which implies it's an isomorphism.

Is there something I'm missing here, because this seems to show that the two sets are isomorphic?

You're more or less right on the money. You've proven that if there is an onto homomorphism, then there is an isomorphism. But the groups cannot be isomorphic (for instance, one has a cyclic subgroup of order $8$). Therefore, by contradiction, there can be no onto homomorphism.
The groups are the same size so any surjective map is injective. What is the order of $(1,0)$ in $\mathbb{Z}_8\oplus\mathbb{Z}_2$? Is there one of those in $\mathbb{Z}_4\oplus\mathbb{Z}_4$?
• how did you think of $(1,0)$? and how do we know for sure there is no element of order 8 in $Z_4 \oplus Z_4$ – So Lo Nov 24 '17 at 19:23