Show N is a normal subgroup and G/N has finite element

In summary: So, if c^n = e, then both c and cN have finite order and thus must be equal. Is that correct?Yes, that is correct. If both c and cN have finite order, then they must be equal since they are both elements of N. This shows that cN is the identity element in G/N, and since the identity element is unique, it follows that cN = N.
  • #1
Robb
225
8
Homework Statement
Let G be an abelian group. Show that the elements of finite order in G form a
normal subgroup N, and that the only element of finite order in G/N is the identity.
Relevant Equations
N/A
Clearly e ∈ N. If a, b ∈ N, say ##a^k = b^l = e##, for some k,l ∈ N, then ##(ab)^{kl} = (a^k )^l (b^l )^k = e^l e^k = e##; thus, ab ∈ N. Also, ##|a|=|a^{−1}|##, so ##a^{−1}## ∈ N. Thus, N is a subgroup. As G is abelian, it is normal. Take any c ∈ G. If, for some n ∈ N, we have ##(cN)^n = eN##, then ##c^n## ∈ N; that is, ##c^n## has finite order, so ##c^{nm} = e## for some m ∈ N. In other words, c ∈ N, so cN = eN.

This is the answer from the back of the book and I am trying to dissect it to understand it. What I don't understand is the part about the only element of finite order in G/N is the identity. So, how do we know ##c^n = e##? How do we know this is the only element of finite order? Why is m necessary (as an exponent to c) if ##c^n = e##? Any help would be greatly appreciated!
 
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  • #2
You assume that ##cN\in G/N## has finite order ##n##. This means ##(cN)^n=c^nN## is the identity coset ##N##. Hence ##c^n\in N##. This implies ##c^n## has finite order, and so ##c## also has finite order. By definition, this gives us ##c\in N##, and ##cN=N## is the identity coset. This means that the identity element ##N\in G/N## is the only element of finite order.
 
  • #3
Infrared said:
You assume that ##cN\in G/N## has finite order ##n##. This means ##(cN)^n=c^nN## is the identity coset ##N##. Hence ##c^n\in N##. This implies ##c^n## has finite order, and so ##c## also has finite order. By definition, this gives us ##c\in N##, and ##cN=N## is the identity coset. This means that the identity element ##N\in G/N## is the only element of finite order.
Much appreciated! I think I get it.
 

Related to Show N is a normal subgroup and G/N has finite element

1. What is a normal subgroup?

A normal subgroup is a subgroup of a group that is invariant under conjugation by elements of the larger group. This means that for any element in the normal subgroup and any element in the larger group, their product and inverse will also be in the normal subgroup.

2. How do you show that N is a normal subgroup?

To show that N is a normal subgroup, you must prove that for any element n in N and any element g in G, the product and inverse of gn and g^-1ng are also in N. This can be done by using the definition of a normal subgroup and the properties of groups.

3. What does it mean for G/N to have finite element?

If G/N has finite element, it means that the quotient group G/N has a finite number of elements. This is important because it tells us that the quotient group is a finite group, which has different properties and behaviors compared to infinite groups.

4. How does having a normal subgroup and a finite quotient group impact the structure of the original group?

Having a normal subgroup and a finite quotient group can give us information about the structure of the original group. For example, if the original group is finite, then the quotient group will also be finite. Additionally, the size of the normal subgroup can give us information about the number of cosets in the quotient group.

5. Are there any practical applications of studying groups with normal subgroups and finite quotient groups?

Yes, there are many practical applications of studying these types of groups. For example, in physics and chemistry, the concept of symmetry and symmetry groups is essential for understanding the behavior of molecules and crystals. These groups often have normal subgroups and finite quotient groups, making the study of these concepts crucial in these fields.

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