- #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!

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!