Proving the elements of a group are of finite order

In summary, if N is a normal subgroup of a group G and every element of N and G/N has finite order, then every element of G has finite order. This can be proven by showing that for every g in G, there exists an m such that (gN)^m=N, and since N is finite and normal, g^m is also in N for some integer m, thus giving g a finite order. This is possible because of the properties of normal subgroups and cosets.
  • #1
quasar_4
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Homework Statement



If N is a normal subgroup of a group G and every element of N and G/N has finite order, prove that every element of G has finite order.

Homework Equations



Um, a finite group element is given such that g^k = e (identity) for some integer k.
We also know that the quotient of a group by a group is itself a group, with cosets as elements. If G/N is the quotient space, N is the identity element.

The Attempt at a Solution



The elements of N, call them n, have some finite order, so for some integer k we can say that n^k=e (the identity). The elements of the quotient group, G/N are the distinct cosets of N in G, so if they have some finite order m, then (Ng)^m=e=Ne (because N is the identity in the quotient group).

I was hoping to do something like saying

(ng)^m = ne ==> (n^m)*(g^m)= ne ==> n^(m-1)*g^m=e ==> somehow g^something =e.


First of all, I am not sure if this is general enough to prove this for ALL cosets of G/N. Second, can I use little n instead of big N (we usually denote the cosets Ng. I think I can do the little one, though, because we're taking an element of N and operating on the left by some group element g). I don't know what else to do! :confused:
 
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  • #2
ooh, wait. Maybe someone can just tell me if this works... I am thinking that maybe

(Ng)^m = N <==> (ng)^m=n. If that is true, then I can say (ng)^m = n ==> n^(m-1)*g^(m)=e ==> (n^(m-1)*g^(m))^k=e^k=e ==> (n^k)^(m-1)*g^(km))=e ==> g^km = e ==> g has finite order.

Does that do it? Does that prove it for all g? Was my first assertion even valid?
 
  • #3
Not quite. Try this. G/N has finite order. So for every g there is an m such that (gN)^m=N. (gN)^m=(g^m)N. If (g^m)N=N then g^m is in N. Can you fill in any missing arguments?
 
  • #4
Oh, that's very cool... I think we'd about be done then, because we already are given the fact that the elements of N are finite as well... So what mechanism is it that allows us to take the m inside the (gN) and just apply it to g? That's the part I am having trouble seeing.
 
  • #5
An element of (gN)^m can be written as g*n1*g*n2*...*g*nm where the n's are in N, right? Since N is normal gN=Ng. So every g*n with n in N can also be written as n'*g where n' is a possibly different element of N. This let's you collect all of the g's together with the rest of the product being a product of n's. The same reason g1*N*g2*N=g1*g2*N.
 

Related to Proving the elements of a group are of finite order

What is a group?

A group is a mathematical structure consisting of a set of elements and an operation that combines any two elements to produce a third element. The operation must be associative, have an identity element, and every element must have an inverse.

What does it mean for elements of a group to have finite order?

The order of an element in a group is the number of times the element must be combined with itself using the group operation to get the identity element. Finite order means that the element can only be combined a finite number of times before reaching the identity element.

How do you prove that the elements of a group are of finite order?

To prove that the elements of a group are of finite order, you must show that each element has a finite order. This can be done by showing that each element can only be combined with itself a finite number of times before reaching the identity element.

Why is it important to prove the elements of a group are of finite order?

Proving the elements of a group are of finite order is important because it helps us understand the structure and properties of the group. It also allows us to make conclusions about the behavior of the group and its elements.

What are some common methods used to prove the elements of a group are of finite order?

One common method is by using the definition of a group and showing that each element has a finite order. Another method is by using the fundamental theorem of cyclic groups, which states that every finite subgroup of a cyclic group is itself a cyclic group.

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