Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)

share|improve this question
6  
Let $f$ be such a homomorphism, what could be the value of $f(1)$ ? –  Bebop Sep 22 '12 at 10:31
2  
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. –  rschwieb Sep 22 '12 at 13:36

2 Answers 2

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.

share|improve this answer
    
why $1$ map to an idempotent element? –  Une Femme Douce May 14 '13 at 10:10
1  
@Tsotsi $\phi(1)\cdot\phi(1)=\phi(1\cdot 1)=\phi(1)$. –  rschwieb May 14 '13 at 11:02
    
okay, and why "...that leaves four options for $1$ to map to"? –  Une Femme Douce May 14 '13 at 11:20
    
@Tsotsi ...because there are only four elements of $\Bbb Z_{28}$ which have orders dividing 12. –  rschwieb May 14 '13 at 11:22
1  
@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. –  rschwieb May 24 '13 at 10:33

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 $<a>_n$ 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 $y<n$. It means that $a$ couldn't be a generator. Therefore if $a^q$ is a generator then $q$ and $n$ are relatively prime.

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: $gcd(12,28)=4$

share|improve this answer
3  
This is almost completely irrelevant: the question is asking about ring homomorphisms, not group homomorphisms. –  Zhen Lin Sep 22 '12 at 12:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.