# no torsion elements in a specific finitely presented group?

I'm trying to prove, or to find a counter example, of the following: Assume we have a group $G$, generated by $x_1,...,x_n$ such that $G/G' \cong \mathbb{Z}^n$. Let $G_1 \doteq G/\langle x_1 = 1 \rangle$. We know that $G_1$ is isomorphic to a direct sum of a free abelian group and free groups, i.e. $G_1 \cong \mathbb{Z}^m + \mathbb{F}_{r_1} + ... + \mathbb{F}_{r_k}$ for some $m,r_1,...,r_k \geq 0$. In particular, we know that there are no torsion elements in $G_1$.

Does this mean that $G$ also does not have torsion elements?

Edit: In $G$, the only allowed relations between the generators are $[x_i,x_j]=1$ and $[x_i,x_jx_k]=1$ (where $j \neq k$).

Thanks, Alex

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Your group is torsion-free. The only torsion could live in $G'$, so we would need an element $z$, in the free group on the $x_i$, which lived in the commutator subgroup, and which had a proper power that was in the relator subgroup. But your relators are all basically primitive (generators for the commutator subgroup), so this doesn't happen. Note that if $[x_i,x_j]=1$, then $[x_i,x_jx_k]=1$ is equivalent to $[x_i,x_k]=1$. – user641 Jan 27 '13 at 18:34
thanks! so in fact the data on $G_1$ is redundant here, right? However, I should note that $G$ can be with only the relations $[x_1,x_2]=1$ and $[x_3,x_4x_5]=1$ (i.e. there isn't necessarily a relation of the form $[x_1,x_2x_j]=1$ for some $j$. Does this cause any problems? (I don't think so, but just to be sure). – user59751 Jan 27 '13 at 18:43
No, it doesn't cause any problems. And yes the data on $G_1$ is reduntant. All you need is the form for the possible relators you gave. If $F$ is the free group on the $x_i$, then you are concerned with torsion in $[F,F]/R$, where $R$ is the relator subgroup. But $[F,F]$ is generated by $[x_i,x_j]$ and their conjugates. So a relator of the form $[x_i,x_j]$ kills a generator, while a relator of the form $[x_i,x_jx_k]=[x_i,x_k][x_i,x_j]^{x_k}$ identifies two generators. So $G'$ is torsion free. – user641 Jan 27 '13 at 18:53
ok, thanks. so in fact, I guess that even if allow that $G$ will have also relations of the form $[x_i,x_jx_kx_j^{-1}]=1$ (where $i,j,k$ all different), that won't cause any problems either? though I guess that in this case $G'$ will not be free group any more. – user59751 Jan 27 '13 at 19:01
Actually you've got to be careful, and I corrected my comment: $G'$ does not have to be free. To see what can go wrong, consider your group to be generated by $x_1,\ldots,x_4$, with relations $[x_1,x_2]=[x_2,x_3]=[x_3,x_4]=[x_4,x_1]=1$. The commutator is not a free group. – user641 Jan 27 '13 at 19:14

Perhaps I am missing something, but if $G$ is defined by generators and relations as $$G = \langle x_1, x_2, y : [x_1, x_2] = y, [x_1, y] = [x_2, y] = y^2 = 1 \rangle,$$ then $G/G' \cong \mathbf{Z}^2$, and $G_{1} \cong \mathbf{Z}$, and $G$ has torsion.
Do you allow $j = k$ in the second type of relations? I'm not sure it makes a difference, but just to be sure. – Andreas Caranti Jan 26 '13 at 17:24