Torsion in the Free Abelian Group

**Swlabr** · May 6th 2010, 02:03 AM

Let $F$ be a free group (not necessarily finitely generated). I am trying to prove that $[F, F]$ has infinite index, and I have, so far, shown that this is equivalent to proving that $F^{\prime} = F/[F, F] = <x_1, \ldots, x_n, \ldots : [x_i, x_j]>$ contains no torsion. Which is obvious.

Except...I can't seem to prove it...

**TheArtofSymmetry** · May 6th 2010, 04:13 AM

Originally Posted by Swlabr

Let $F$ be a free group (not necessarily finitely generated). I am trying to prove that $[F, F]$ has infinite index, and I have, so far, shown that this is equivalent to proving that $F^{\prime} = F/[F, F] = <x_1, \ldots, x_n, \ldots : [x_i, x_j]>$ contains no torsion. Which is obvious.

Except...I can't seem to prove it...

$F^{\prime}$ is simply a free abelian group. A free abelian group is the internal direct sum of a family of infiinite cyclic subgroups, so it is torsion-free.

**Swlabr** · May 6th 2010, 04:26 AM

Originally Posted by TheArtofSymmetry

$F^{\prime}$ is simply a free abelian group. A free abelian group is the internal direct sum of a family of infiinite cyclic subgroups, so it is torsion-free.

Yeah, that makes sense. I knew it was obvious!

Thanks.

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