# Homomorphism and group element orders

I return to some old questions that left me puzzled in the past but was none the wiser taking a second look.

Question: Prove that no homomorphism from $\mathbb{Z}_{8}\oplus \mathbb{Z}_{2}$ to $\mathbb{Z}_{4}\oplus \mathbb{Z}_{4}$ exists.

The order of the elements in $\mathbb{Z}_{8}\oplus \mathbb{Z}_{2}$ are 1,2,8 and 4. The order of the elements in $\mathbb{Z}_{4}\oplus \mathbb{Z}_{4}$ are 4. Now, homomorphism does not require preservation of group element orders. One of the condition is that the order of an element g under a homomorphism map divides the order of the element g. However, looking a look at some solutions, it is mentioned that because group element orders are not preserved in this case, the map is not isomorphic and therefore no homomorphism. This doesn't come across as very sensible.

Any help is appreciated.

• The question was this one, and you forgot that it asks for a surjective homomorphism. Otherwise it is possible. Take $\phi=0$. Commented Apr 24, 2017 at 14:51
• @DietrichBurde Surjection is not a necessary condition for the definition of a homomorphism, or is it? The definition demands only that group operation is preserved Commented Apr 24, 2017 at 14:56
• Further, in the link you've provided. All that it says is that the map is not an isomorphism since isomorphism demands that the order of an element be preserved but clearly group element order is not preserved here. So all that can be said is that the map is not an isomorphism. Commented Apr 24, 2017 at 15:00
• Then which "old question" do you mean? Could you give a link? Your present question as written is wrong. There is a homomorphism between these two groups. Commented Apr 24, 2017 at 15:02
• Yes, it exists: take $\phi(x,y)=0$ for all $x,y$. And the real question asks: Prove that no onto homomorphism exists. Commented Apr 24, 2017 at 15:06

Your question is not correct as Dietrich Burde mentioned in the comment.

For any groups $G,H$, define $\phi:G\rightarrow H$ by $\phi(g)=1$ for every $g\in G$.
Then $\phi$ is a homomorphism from $G$ to $H$.
This is called the trivial homomorphism.

• And if the groups are written additively, it is the zero homomorphism. Commented Apr 24, 2017 at 15:06
• This makes sense now. The author was possibly then asking for the readers to show that no isomorphism exists; for which it can be demos rated. Commented Apr 24, 2017 at 15:07
• Ok what about non trivial homomorphism, as that is not really a map to every element $H$. Good point of technicality though. Commented Apr 24, 2017 at 15:12
• @Mathematicing it hypothetically could have subgroup of $H$ not having order $1$, so I dont think showing now isomorphism would be enough. Commented Apr 24, 2017 at 15:16
• Also can I interpret the direct sum as simply the product group, ie n tuple coming from each group? Commented Apr 24, 2017 at 15:23

Also what about homomorphism from $$\Bbb Z_8 \times \Bbb Z_2 \to \Bbb Z_4 \times \Bbb Z_4$$ given by $\phi(x,y)=(x \ mod \ 4, \ x \ mod \ 4)$?

• The "old question" the OP is referring to, was this one. There the title says onto homomorphism, but the question in the body forgot about the "onto". Of course, there are homomorphisms between the two groups. Commented Apr 24, 2017 at 15:37
• I see [some characters]. Commented Apr 24, 2017 at 15:40