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Simple question (I seem have asked a few like this...)

What is $\mbox{Hom}(\mathbb{Z}/2,\mathbb{Z}/n)$? (for $n \ne 2$)

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So what could it possibly be? How many elements of order two are there in $\mathbb{Z}/n$? – t.b. May 10 '11 at 7:12
@Theo - I am not sure what it could be! (I'm sure I'm missing something obvious). The even $\mathbb{Z}/n$ groups have 1 element of order 2... – Juan S May 10 '11 at 7:28
The questions you likely should be asking is "What is a homomorphism" and "what is a cyclic group". If you are comfortable with the answers to these questions, I can't envisage you not being able to answer the one you posted. – Alex B. May 10 '11 at 7:28
up vote 4 down vote accepted

Try to find the neccessary condition for such a homomorphism first.

If f is a homomorphism from Z/2 to Z/n then clearly f maps the zero element of Z/2 to the zero element of Z/n. Can it map 1 to an arbitrary element of Z/n? No, the mapping should be such that f is a homomorphism. Assuming it is, and $f(1) = x$, clearly $0 = f(0) = f(1+1) = f(1) + f(1) = x + x$ should hold, following which $2x = 0$ in Z/n. So this is a neccessary condition.

It is easy to see that this condition is sufficient as well. So the homomorphisms may be counted in number by counting all such x. This amounts to counting all solutions to the congruence $2x = 0$ in Z/n.

Clearly as $gcd(2,n)|0$ so solutions always exist and they will be gcd(2,n) (which may be either 1 or 2, depending upon whether n is odd or even repectively) in number. So Hom(Z/2,Z/n) will have either 1 or 2 elements. In either case, it will be cyclic because all groups of order 1 and 2 are.

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For the general case proceed similarly. A bijection between the subgroup $\{x\in Z/n:mx=0(mod n)\}$ and $Hom(Z/m,Z/n)$ establishes the cyclic nature of $Hom(Z/m,Z/n)$. – Shahab May 10 '11 at 12:44

More generally $\operatorname{Hom}{(\mathbb{Z}/m,\mathbb{Z}/n)}$ is cyclic of order $\operatorname{gcd}(m,n)$. I strongly recommend you to try to prove this on your own. In case of emergency, google for hom z nz z mz.

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thanks, I got it just after Alex's hint (the downside of - sometimes too quick to post questions) – Juan S May 10 '11 at 7:35
@Qwirk This is not necessarily just a phenomenon of online sites. Later in the common room you will find yourself in a similar situation, when it is easy to ask an expert a question, since he is already in front of you. But as a general rule of thumb for the future (should that future involve mathematics on a professional level), I recommend this: to display ignorance in mathematics is great. To display laziness is very bad. – Alex B. May 10 '11 at 7:46
@Alex: I couldn't agree more. – t.b. May 10 '11 at 7:47
@Alex: Well put! – Juan S May 10 '11 at 8:12

If $f:G\rightarrow H$ is a group homomorphism, then $\text{ord}(f(g))\mid\text{ord}(g)$ for all $g\in G$ because $g^n=e_G$ implies $f(g)^n=f(g^n)=f(e_G)=e_H$.

Thus, if $f:\mathbb{Z}/(2)\rightarrow H$ is a homomorphism, we know that $f(0+(2))=e_H$ because $f$ is a homomorphism, and $f(1+(2))$ must be an element of order dividing 2 in $H$. In fact, given any element $h\in H$ of order dividing 2, we can define a homomorphism $f:\mathbb{Z}/(2)\rightarrow H$ by $f(0+(2))=e_H$ and $f(1+(2))=h$.

Thus, for any group $H$, the homomorphisms $f:\mathbb{Z}/(2)\rightarrow H$ are in bijection with the elements of order dividing 2 in $H$. These are the elements of order 2, along with the identity $e_H$ (which is the only element of order 1, obviously). This explains Theo's suggestion above.

Hint: An element $a+(n)$ of $\mathbb{Z}/(n)$ is of order 2 when $a+(n)\neq 0+(n)$, but $$2(a+(n))=2a+(n)=0+(n),$$ or in other words, $a\not\equiv0\bmod n$ but $$2a\equiv 0\bmod n.$$ For which $n$ does such an $a$ exist?

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