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How do I find all group homomorphisms from $\mathbb{Z_2} \times \mathbb{Z_2}$ to $\mathbb{Z_6}$?

$\mathbb{Z_2} \times \mathbb{Z_2}$ $= \{ (0,0), (0,1),(1,0),(1,1) \}$

and

$\mathbb{Z_6} = \{0, 1, 2, 3, 4, 5 \}$.

In $\mathbb{Z_2} \times \mathbb{Z_2}$ : $|(0,0)| = 1, |(0,1)| = 2, |(1,0)| = 2,$ and $|(1,1)|=2.$

In $\mathbb{Z_6}$ : $|0| = 1, |1| = 6, |2| = 3, |3| = 2, |4| = 3,$ and $|5| = 6.$

In the case of an isomorphism, elements of a group can only be mapped to elements of the other group of the same order. I understand that in a homomorphism the same rule does not apply.

What are the options for mapping elements of $\mathbb{Z_2} \times \mathbb{Z_2}$ to elements in $\mathbb{Z_6}$? Can elements in the domain be mapped to elements in the codomain that divide such an element?

The trivial homomorphism is just the map that takes $(0,0)$ to $0$.

I also understand that $f(a*b)= f(a) \bullet f(b)$, i.e., an operation preserving map.

Thank you for your time and help.

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  • $\begingroup$ The order of the image of an element by a homomorphism divides the order of the element. Hence the generators $(1,0)$ and $(0,1)$ must be sent by a homomorphism to an element in $\{0,3\}$. $\endgroup$ – Nex Nov 5 '15 at 5:17
  • $\begingroup$ @Nex So where can (1,0) and (0,1) be mapped to in $\mathbb{Z_6}$? $\endgroup$ – mathamphetamines Nov 5 '15 at 5:23
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    $\begingroup$ I thought I explained that in the second sentence? $(1,0)$ and $(0,1)$ must be mapped to either $0$ or $3$ (try all posibilities). $\endgroup$ – Nex Nov 5 '15 at 5:40
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Here's a useful tool: Let $G$ and $G'$ be groups, let $f \colon G \to G'$ be a group homomorphism. If $x \in G$ and $\vert x \vert = n$, then $$f(x)^n = f(x^n) = f(e) = e,$$ and so $\vert f(x) \vert$ must divide $n$. Using this will cut down significantly in your search for homomorphisms.

Since every nonidentity element of $\mathbb{Z}_2 \times \mathbb{Z}_2$ has order $2$, they must all get sent to an element in $\mathbb{Z}_6$ with order $1$ or $2$.

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    $\begingroup$ An element of order dividing 2, as you note. No homomorphism can send every nonidentity element to $3 \in \mathbb{Z}_{6}$. Such a map would have trivial kernel, but this can't be, as all nonidentity elements map to the same element of $\mathbb{Z}_{6}$. $\endgroup$ – Alex Wertheim Nov 5 '15 at 6:04
  • $\begingroup$ Your answer is currently wrong please fix it. Please replace "to an element of order two" by something equivalent to "to an element whose order divides two". $\endgroup$ – Nex Nov 5 '15 at 17:57
  • $\begingroup$ My mistake, thanks! $\endgroup$ – Ethan Alwaise Nov 6 '15 at 1:06

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