Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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

2 Answers 2

Perhaps I am missing something, but if $G$ is defined by generators and relations as \begin{equation} G = \langle x_1, x_2, y : [x_1, x_2] = y, [x_1, y] = [x_2, y] = y^2 = 1 \rangle, \end{equation} then $G/G' \cong \mathbf{Z}^2$, and $G_{1} \cong \mathbf{Z}$, and $G$ has torsion.

share|improve this answer
    
yes, of course, let me edit the question... I forgot to put one and important) constraint on the relations. –  user59751 Jan 26 '13 at 17:17
    
Ah, ok, looking forward to the revised version. –  Andreas Caranti Jan 26 '13 at 17:19
    
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
    
no, it is not allowed (added that, thanks). –  user59751 Jan 26 '13 at 17:25

I don't know about torsion elements but your quotient seems a bit off to me. Setting one of the generators to the identity simply gives you a free group on n-1 generators.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.