# Torsion-free group

• Jul 27th 2011, 05:42 PM
deniselim17
Torsion-free group
My question is simple.
Subgroup of a torsion-free group is torsion-free again.
True or false?

A group $\displaystyle G$ is called torsion-free if every element in $\displaystyle G$ has infinite order except the identity.

We know that infinite group has infinite number of subgroups.
But are the subgroups of infinite order too?

So, for my question here, is it true that all the elements in any subgroup of torsion-free group $\displaystyle G$ has infinite order?
• Jul 28th 2011, 12:44 AM
Swlabr
Re: Torsion-free group
Quote:

Originally Posted by deniselim17
My question is simple.
Subgroup of a torsion-free group is torsion-free again.
True or false?

A group $\displaystyle G$ is called torsion-free if every element in $\displaystyle G$ has infinite order except the identity.

We know that infinite group has infinite number of subgroups.
But are the subgroups of infinite order too?

So, for my question here, is it true that all the elements in any subgroup of torsion-free group $\displaystyle G$ has infinite order?

Yes. I mean, if you had $\displaystyle h\in H\leq G$ with $\displaystyle h^n=1$ then $\displaystyle h^n=1$ in both $\displaystyle G$ and $\displaystyle H$. I amn't even sure how to explain this...it is kinda fundamental...! Basically, if $\displaystyle H$ is a subgroup, $\displaystyle G$ contains a copy of it - everything that happens in $\displaystyle H$ happens somewhere in $\displaystyle G$!

That said, it isn't true that homomorphic images of torsion-free groups are torsion-free. This is obvious if you know about free groups, but if you do not then take the integers under addition. It is an interesting exercise to prove that every homomorphic image of $\displaystyle \mathbb{Z}$ is finite.