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

[Robb]

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!

 

[Infrared]

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.

 

[Robb]

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.