I am trying to understand homomorphisms from $\displaystyle Z_n $ to $\displaystyle Z_k $ and despite some help from earlier posts am still struggling.

I am thus reading Beachy and Blair: Abstract Algebra on group homomorphisms and am trying to follow Example 3.7.7 on page 156 which seeks to determine all homomorphisms from $\displaystyle Z_n $ to $\displaystyle Z_k $ (see relevant page attached). I can follow this example but am having trouble with a particular step.

I will present their argument down to the problematic step - and I will make the example more concrete by seeking to determine all homomorphisms $\displaystyle \phi $: $\displaystyle Z_6 $ $\displaystyle \rightarrow $ $\displaystyle Z_{10} $

In terms of my particular example the argument of Beachy and Blair is as follows:

================================================== ====

Consider $\displaystyle \phi $: $\displaystyle Z_6 $ $\displaystyle \rightarrow $$\displaystyle Z_{10} $

Any such homomorphism is completely determined by $\displaystyle \phi $ ($\displaystyle [1]_6$), and this must be an element $\displaystyle [m]_{10}$ of $\displaystyle Z_{10} $ whose order is a divisor of 6.

In an abelian group with the operation denoted additively we have that if o(a) | n if and only if n.a = 0.

Applying this result to $\displaystyle [m]_{10}$ in $\displaystyle Z_{10} $ we have o($\displaystyle [m]_{10}$) | 6 if and only if n. $\displaystyle [m]_{10}$ = $\displaystyle [0]_{10}$(Beachy and Blair ask us to compare with Exercise 11 of Section 2.1 - see other attached sheet)which happens if and only if 10 | 6m

etc etc - see attached sheet

================================================== ==

My question is how does the step that concludes " which happens if and only if 10 | 6m" follow ie how does this last step of the argument follow?

I would be grateful for any help!

Peter