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

Robb
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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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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.
 
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.
 
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