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The number of non-trivial ring homomorphisms from $\mathbb{Z}_{12}$ to $\mathbb{Z}_{28}$ is (Options: a.1 b.3 c.4 d.7)

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    Let $f$ be such a homomorphism, what could be the value of $f(1)$ ?2012-09-22
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    A ring homomorphism preserving identity would have to send $\phi(0)=\phi(12)=12\neq 0$, so I'm beginning to suspect that we're talking about groups after all, or the ring homomorphisms do not have to preserve identity.2012-09-22

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To be a mere group homomorphism, then the order of the image of 1 will have to divide both 12 and 28. So that leaves four options for 1 to map to.

If it has to be a ring homomorphism, then 1 must map to an idempotent element of $\mathbb{Z}_{28}$. Only two of the four previous possibilities are idempotent.

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    why $1$ map to an idempotent element?2013-05-14
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    @Tsotsi $\phi(1)\cdot\phi(1)=\phi(1\cdot 1)=\phi(1)$.2013-05-14
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    okay, and why "...that leaves four options for $1$ to map to"?2013-05-14
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    @Tsotsi ...because there are only four elements of $\Bbb Z_{28}$ which have orders dividing 12.2013-05-14
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    namely $O(7)=4,O(0)=1,O(14)=2,O(17)=6$, In which $0$ is the only idempotent element, do want to mean that?2013-05-14
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    @Tsotsi Don't combine the two different hints above. Being an idempotent has nothing to do with the group map in the first paragraph.2013-05-14
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    so then clearly state what is the answer and where I am going wrong2013-05-14
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    @Tsotsi I didn't see you "go wrong" besides combining the two hints :) The answer is that you can map $1\mapsto x$ where $x\in \{0,7, 14,21\}$, those all induce group homomorphisms, and nothing else is possible :) If you want a *ring* homomorphism then you're forced to map $1\mapsto 0$, (because the *ring* homomorphism needs to send 1 to an idempotent.)2013-05-14
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    @Tsotsi It isn't possible for $17$ to have order $6$ since $6$ doesn't divide $28$. Since it's coprime to $28$ it has order $28$.2013-05-14
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    21 is also an idempotent element2013-05-24
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    @TaxiDriver :o So it is! I have no idea why I overlooked that before! I will have to revise that omission, and I thank you for pointing it out.2013-05-24
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I'm, going to proove more general fact: the order of the group $Hom(\mathbb Z_m, \mathbb Z_n)$ (i.e the group of homomorphisms from $\mathbb Z_m$ to $\mathbb Z_n$) is $gcd(m,n)$.

It's obvious that the order of the image of any homomorphism from $\mathbb Z_m$ to $\mathbb Z_n$ must devide both $m$ and $n$. Let us notice that for any $d$ that divides $n$ there exists a unique subgroup $H$ in $\mathbb Z_n$ which order is $d$ (if $n=dq$ then $H=\{0, d, 2d, ... , d(q-1)\}$).

There is also a simple fact that the number of generators of finite cyclic group $_n$ of order $n$ is $\phi(n)$, where $\phi$ is Euler's totient function (by definition $\phi(n)$ is an arithmetic function that counts the number of positive integers less than or equal to $n$ that are relatively prime to $n$). Now I'm going to prove it.

At first let's show that if $a^q$ is a generator then $gcd(q,n)=1$. Assume that $q$ and $n$ aren’t relatively prime. Therefore $q = kx$, and $n = ky$ for some integers $x$ and $y$. This means that $a^{qy} = a^{kxy} = a^{xn}=1$. So the order of $a^q$ is $y$. But $y

Now we want to show that if $gcd(q,n)=1$, then $a^q$ is a generator. More pricisely, we need to proove that if $(a^q)^s = 1$, then $s = xn$ for some integer $x$. It's obvious that $qs=xn$ for some integer $n$ (because $(a^q)^s = 1$ and our group has order $n$). But $gcd(q,n)=1$, so it's easy to see that $n$ devides $s$.

It remains to observe that given a homomorphism from $\mathbb Z_m$ to subgroup $H$ in $\mathbb Z_n$ means to set the map of $1$ to one of generators of $\mathbb Z_n$. Therefore the number of homomorpisms is $\sum_{k|gcd(m,n)} \phi(k)$ which equals $gcd(m,n).$

So the answer to your question: $gcd(12,28)=4$

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    This is almost completely irrelevant: the question is asking about _ring_ homomorphisms, not group homomorphisms.2012-09-22