1
$\begingroup$

I'm having difficulty in finding all homomorphims when the numbers are very high. First I'll do a example of a not so high. For example:

Finding all homomorphisms of $f:\mathbb{Z}_6\to\mathbb{Z}_{12}$

If $f$ is a homomorphism, $f$ is determinated by $f(1_6)$. So, $f(1_6)$ can be any element of $\mathbb{Z}_{12}$ whose order divides $6$. Orders of the elements in $\mathbb{Z}_{12}$:

  • $0_{12}$ = $id$ $\to$ Order=$1$
  • $1_{12}$ $\to$ Order =$12$
  • $2_{12} \to$ Order=$6$
  • $3_{12} \to$ Order=$4$
  • $4_{12} \to$ Order=$3$
  • $5_{12} \to$ Order=$12$
  • $6_{12} \to$ Order=$2$
  • $7_{12} \to$ Order=$12$
  • $8_{12} \to$ Order=$3$
  • $9_{12} \to$ Order=$4$
  • $10_{12} \to$ Order=$6$
  • $11_{12} \to$ Order=$12$

We want the elements whose order divides $6$, so $f(1_6)=\{0_{12},2_{12},4_{12},6_{12},8_{12},10_{12}\}$

It took some time, but I managed to solve it. But, for this example:

Finding all homomorphisms of $f:\mathbb{Z}_{12}\to\mathbb{Z}_{60}$

What do I do with such high numbers? Finding the order of all of the elements in $\mathbb{Z}_{60}$ takes alot of time. Is there any similiar method for high numbers?

Thanks in advance.

2 Answers 2

1

The number of group homomorphisms $\phi:\mathbb{Z}/m\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ is given by $d=(m,n)$, where $(m,n)$ is the gcd of $m$ and $n$.

Indeed, $\phi$ is determined by $k=\phi(1)$, and we must have $mk\equiv 0$ (mod $n$), or $\frac{m}{d}k\equiv 0$ (mod $\frac{n}{d}$). Since $\frac{m}{d}$ and $\frac{n}{d}$ are coprime, this implies that $k$ must be divisible by $\frac{n}{d}$, hence there are $\frac{n}{\frac{n}{d}}=d$ possibilities for $k$.

  • 0
    but for rings ? Sorry, someone sent me on this page, but it was not my original question see here : http://math.stackexchange.com/questions/2167225/ring-homomorphism-of-mathbb-z-k-mathbb-z-to-mathbb-z-m-mathbb-z – 2017-03-01
0

The order of an element $a$ in $\Bbb Z_{60}$ is $\frac {60}{\gcd(a,60)}$ so you can look at multiples of the factors of $60$. So $30$ has order $2.$ $20$ and $40$ have order $3$, and so on. To find an element of order $12$ you need an element divisible by $5$ but not $2$ or $3$, so $5,25,35,55$