# Finding a normal subgroup H of Zmn of order m

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1. Mar 4, 2015

### Rick Strut

1. The problem statement, all variables and given/known data
Find a normal subgroup H of Zmn of order m where m and n are positive integers. Show that H is isomorphic to Zm.

2. Relevant equations

3. The attempt at a solution
I am honestly not even sure where to start. My initial thoughts were if Zmn was isomorphic to Zm x Zn then I could find a subgroup H from that group. However, I discovered that Zmn is isomorphic to Zm x Zn but the converse is not true. Any help would be appreciated.

Edit: If Zmn is cyclic has an element of order mn say x. Then nx has order m. Let H=⟨nx⟩.
Now I just need to show that H is isomorphic to Zm, by constructing an isomorphism.

Last edited: Mar 4, 2015
2. Mar 4, 2015

### Dick

You are just working with numbers mod mn here. Just put x=1. Think about the set of numbers {0,n,2n,3n,...,(m-1)n} like in your edit. Can't you see a correspondence with {0,1,2,3,...,(m-1)}?

Last edited: Mar 4, 2015
3. Mar 4, 2015

### Stephen Tashi

What do you mean by that? Shouldn't you make an "If...then...." statement in order to speak of a converse?

Instead of thinking about isomorphisms, you could think about a homomorphism and the "first isomorphism theorem" for groups. http://en.wikipedia.org/wiki/Isomorphism_theorem

4. Mar 4, 2015

### Rick Strut

I do. However, I don't see how to construct a function based on this. I apologize for not being clear but the goal is to construct a bijective homorphic function. Then I can conclude that they are isomorphic.

5. Mar 4, 2015

### Dick

I would say try $\phi(kn \pmod {nm})=k \pmod m$ just from looking at it. Try to prove $\phi$ is an isomorphism.

Last edited: Mar 4, 2015