abelian groups

**mpryal** · April 23rd 2009, 10:54 AM

Let G be an abelian group (possibly infinite and let the set T={a in G|a^m=e, m>1 depending on a}. Prove that:

a) T is a subgroup of G.
b) G/T has no element - other than its identity element - of finite order.

Please show steps. This section confuses me so much.

Thank you.

**Gamma** · April 23rd 2009, 05:34 PM

All you gotta do to show something is a subgroup is show it is closed under the group operation and closed under inverses (the rest is inherited from the mother group).

This is the torsion subgroup. Let $a,b\in T$ . Then $\exists m, n \in \mathbb{N}$ such that $a^m=1=b^n$ . We wish to show $ab\in T$ .
This is particularly easy because G is assumed to be abelian. $(ab)^{mn}=a^{mn}b^{mn}=(a^m)^n(b^n)^m=1^n 1^m=1$
Thus $ab\in T$ as desired.

Let $a\in T \subset G \Rightarrow a^{-1}\in G$ , suppose $a^n=1$ .

$(a^{-1})^n=a^{-n}=(a^n)^{-1}=1^{-1}=1$

Thus $a^{-1}\in T$ as desired
T is a subgroup of G.

You just modded out by everything of finite order by definition of T. G/T cannot have anything other than the coset 1T itself of finite order, think about it. G is abelian so every subgroup is normal so this is a well defined group.

Just a note, if G were finite everything is torsion by LaGrange, so G/T is trivial. The only interesting things are for infinite abelian groups.

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