Finding a normal subgroup H of Zmn of order m

In summary, the conversation discusses finding a normal subgroup H of Zmn of order m and showing its isomorphism to Zm. The approach involves considering the set of numbers {0,n,2n,3n,...,(m-1)n} and finding a correspondence with {0,1,2,3,...,(m-1)}. The use of a homomorphism and the "first isomorphism theorem" is suggested to prove the isomorphism. A possible function to construct is ##\phi(kn \pmod {nm})=k \pmod m##.
  • #1
Rick Strut
2
0

Homework Statement


Find a normal subgroup H of Zmn of order m where m and n are positive integers. Show that H is isomorphic to Zm.

Homework Equations

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.
 
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  • #2
Rick Strut said:

Homework Statement


Find a normal subgroup H of Zmn of order m where m and n are positive integers. Show that H is isomorphic to Zm.

Homework Equations

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.

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)}?
 
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  • #3
Rick Strut said:
I discovered that Zmn is isomorphic to Zm x Zn but the converse is not true.
What do you mean by that? Shouldn't you make an "If...then..." statement in order to speak of a converse?

Now I just need to show that H is isomorphic to Zm, by constructing an isomorphism.
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
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
Rick Strut said:
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.

I would say try ##\phi(kn \pmod {nm})=k \pmod m## just from looking at it. Try to prove ##\phi## is an isomorphism.
 
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1. What does it mean to find a normal subgroup of Zmn?

Finding a normal subgroup of Zmn means identifying a subset of the group Zmn that is closed under the group operation, and also has the property that for any element in the subgroup and any element in the larger group, their product will also be in the subgroup.

2. Why is it important to find a normal subgroup of Zmn?

Finding a normal subgroup of Zmn can be useful in many applications, including group theory, abstract algebra, and cryptography. Normal subgroups have special properties that make them easier to work with mathematically, and they can also help to simplify complex problems involving the group Zmn.

3. How do you determine the order of a normal subgroup of Zmn?

The order of a normal subgroup of Zmn is equal to the number of elements it contains. In order to determine the order, you would need to list out all the elements of the subgroup and count them. Keep in mind that the identity element (0 in the case of Zmn) is always included in the subgroup, so the order will be at least 1.

4. Can there be more than one normal subgroup of Zmn with the same order?

Yes, it is possible for there to be multiple normal subgroups of Zmn with the same order. For example, in the group Z6, both the subgroups {0, 2, 4} and {0, 3} have an order of 3 and are normal subgroups.

5. How can I check if a subgroup of Zmn is normal?

There are a few ways to check if a subgroup of Zmn is normal. One way is to use the definition of a normal subgroup, which states that for any element in the subgroup and any element in the larger group, their product must also be in the subgroup. Another way is to use the fact that in Zmn, all subgroups of a given order are conjugate, so if you can find a conjugate of the subgroup that is not equal to the original subgroup, then it is not normal.

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