# [SOLVED] homomorphism issue on Z_m \to Z_n

• May 1st 2009, 07:21 AM
hatsoff
[SOLVED] homomorphism issue on Z_m \to Z_n

Quote:

Write down the formulas for all homomorphisms from $\mathbb{Z}_6$ into $\mathbb{Z}_9$.
Choose a homomorphism $\phi:\mathbb{Z}_6\to\mathbb{Z}_9$ with $a,b\in\{0,1,2,3,4,5\}$. Then $\phi([a+b]_6)=\phi([a]_6)+\phi([b]_6)$. What else can we say about $\phi$? Nothing significant, as far as I can tell.

But my textbook has another story. According to it, there are only

Quote:

3 different homomorphisms, given by the formulas $\phi_3([x]_6)=[3x]_9$, $\phi_6([x]_6)=[6x]_9$, or $\phi_0([x]_6)=[0]_9$, defined for all $[x]_6\in\mathbb{Z}_6$.
First of all, I am not completely certain I understand the notation, here. Is $\phi_n(\alpha)$ simply another way of writing $[\phi(\alpha)]^n$? If so, then by algebra of integers modulo, isn't $\phi_n([x]_6)=\phi([nx]_6)$ true for all $n$? If not, then what does the notation mean?

As you can see, I'm a little bit befuddled by this. I would welcome any assistance.

Thanks!
• May 1st 2009, 01:28 PM
ThePerfectHacker
Quote:

Write down the formulas for all homomorphisms from $\mathbb{Z}_6$ into $\mathbb{Z}_9$.
If $\phi : \mathbb{Z}_6 \to \mathbb{Z}_9$ is a homomorphism then it is completely determined by $\phi( [1]_6)$ since $[1]_6$ generates the group $\mathbb{Z}_6$.
If $\phi([1]_6) = [n]_9$ then $\phi([x]_9) = [nx]_9$.

But if $\phi$ is is homomorphism it means the order of $[1]_6$ must be dividsibly by order of $\phi ([1]_9) = [n]_9$. The order of $[1]_6$ is six. Therefore, the order of $[n]_9$ in $\mathbb{Z}_9$ must be have order $1,2,3,6$. It cannot have order $2,6$ since $2,6\not | 9$ and so only two possibilites remain, that the order of $[n]_9$ must be $1,3$. For order $1$ we get $[n]_9 = [0]_9$, for order $3$ we get $[3]_9,[6]_9$ have order $3$.

The the only possibily homomorphism are:
$\phi ([x]_6) = [0]_9, \phi([x]_6) = [3x]_9, \phi([x]_6) = [6x]_9$.

Now check that these are indeed all homomorphisms.