Find the kernel of morphism $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}$

Question

Let be $n$ and $m$ two integers such that $m\mid n$.

I want to show that there exists an homomorphism onto between $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}$.

I find the homomorphism define by $f(a+n\mathbb{Z})= a+m\mathbb{Z}$. Now i want to find it kernel. So I solve $a+n\mathbb{Z}= m\mathbb{Z}$, but i don't know how to conclude

Is https://math.stackexchange.com/questions/2072746/finding-all-mathbb-z-module-homomorphisms-mathrmhom-mathbb-z-n-mathbb-z any help? — Gerry Myerson, Nov 15 '17 at 11:32

David Reed · Answer 1 · 2017-11-15T11:49:26.053

2

Try writing $m\mathbb{Z} = 0 + m\mathbb{Z}$. For an arbitrary abelian group,written additively, $$a + H =b+ H \iff a-b \in H$$

Also, in the specific case of $\mathbb{Z} / \mathbb{nZ}$,

$$a + n\mathbb{Z} = b + n\mathbb{Z} \iff n | \space (a-b) \iff a = b + nk \text{ for some } k \in \mathbb{Z}$$

edited Nov 15 '17 at 11:49

answered Nov 15 '17 at 11:43

David Reed

3,335

i agree your post but i don't understand where do you want to go – Farouk Deutsch Nov 15 '17 at 12:17
Since $m|n$, $x=y \ mod n \implies x= y \ mod m$ – David Reed Nov 15 '17 at 12:51

score 1 · Accepted Answer · answered Nov 15 '17 at 13:10

Since $m \mid n$, there exist an integer, M, such that $n = Mm$.

If $a+n\mathbb{Z}= m\mathbb{Z}$, then there are integers, $u$ and $v$, such that

\begin{align} a + un &= vm \\ a &= vm - un \\ a &= vm - uMm \\ a &= (v-uM)m \\ m &\mid a \end{align}

It follows that $\ker f = \{ 0 + \mathbb Z_n, m + \mathbb Z_n, 2m + \mathbb Z_n, \dots, (M-1)m + \mathbb Z_n \}$

Find the kernel of morphism $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}$

2 Answers2