3

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?

Oliver G
  • 4,792
  • Just for the users who would be interested, here is one of your related questions: http://math.stackexchange.com/questions/1902489 – Watson Aug 24 '16 at 22:18

2 Answers2

5

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.

Note that your argument that an onto homomorphism must be injective, while more or less correct, can be done a bit easier. Any onto function between finite sets of the same size must be injective. This is specifically true for any function that happens to be a homomorphism.

Arthur
  • 199,419
1

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 $\mathbb{Z}_4\oplus\mathbb{Z}_4$?

ItsNotMeItsYou
  • 836
  • 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