I know how to determine all homomorphisms from $\def\Z{\Bbb Z}\Z_n$ to $\Z_m$ (and to itself, naturally), but I can't seem to find an approach towards determining all homomorphisms from $\Z_n \times \Z_m$ to itself. For example, from $\Z_2\times\Z_2$ to itself.
Determining all homomorphisms from $\Bbb Z_n \times \Bbb Z_m$ to itself
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group-theory
group-homomorphism
1 Answers
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You can simply send the generator of $\def\Z{\Bbb Z}\Z/n\Z$ to any element $x$ satisfying $nx=0$, and the generator of $\def\Z{\Bbb Z}\Z/m\Z$ to any element $y$ satisfying $my=0$. Then of course $(a,b)$ maps to $ax+by$. The sets of allowed values for $x,y$ are easy to determine, but depend somewhat on possible common divisors of $n$ and $m$.
For instance for $(\Z/2\Z)\times(\Z/2\Z)$, all elements satisfy $2x=0$, so you have $4^2=16$ different group endomorphisms.
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0Are you referring to Z/Zn as to a quotient group? If so, that's mauly not be what I'm searching for. – 2017-01-08
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0So if not a quotient group, what _do_ you mean by $\Z_n$, the [$n$-adic integers](https://en.wikipedia.org/wiki/P-adic_number)? If that is the case you had probably better say it clearly in your question, because most people here will read $\Z_n$ as "the integers modulo $n$" instead (as indeed I did). – 2017-01-08