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Let $G$ be the group given by the presentation $\langle x , y , z : x^2 , y^3 , (xyz)^4 \rangle$. I would like to write $G/[G , G]$ as a direct product of cyclic groups, where $[G , G]$ is the commutator subgroup of $G$.

I am familiar with writing (finitely generated) abelian groups as a product of cyclic groups using Smith normal forms, but the free group aspect of this problem is throwing me off (which I believe is the point). I am aware that the free group on a set $S$ modulo its commutator subgroup is isomorphic to $\mathbb{Z}^{|S|}$, but I am unsure of how the additional relations (such as those in the problem in question) impact things.

Obviously $G/[G , G]$ is a finitely generated abelian group, but is $G/[G , G]$ isomorphic to the abelian group, $A$, (written additively) generated by elements $a , b, c \in A$ subject to the relations $2a = 3b = 4(a + b + c) = 0$? If this is the case, then I am completely comfortable finishing off the problem and actually writing $G/[G , G]$ as a direct product of cyclic groups, but I would like some more justification as to why. If this is not the case, then what is the correct way to go about this problem?

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    $\begingroup$ Yes; $G/[G,G]$ has the presentation you start with together with the relations $[x,y]=[x,z]=[y,z]=1$. So it reduces to the conditions you give. $\endgroup$ Jul 12 '21 at 16:50
  • $\begingroup$ Can you elaborate on what type of "justification as to why" you would be looking for? $\endgroup$ Jul 12 '21 at 16:58
  • $\begingroup$ @ArturoMagidin I think this is what I am looking for and along the lines of what I was thinking in the first place. The map that sends generators of a free group on $S$ to generators of $\mathbb{Z}^{|S|}$ induces an isomorphism between the free group on $S$ modulo its commutator and some relations $R$ onto $\mathbb{Z}^{|S|}$ modulo the relations corresponding to $R$. $\endgroup$
    – Oiler
    Jul 12 '21 at 18:27
  • $\begingroup$ Your comment does not seem to contain a question, so I am at a loss as to what it is you are looking for. $\endgroup$ Jul 12 '21 at 18:30
  • $\begingroup$ @ArturoMagidin I had the same thought, but I guess I wanted more details as to why “reduces to the conditions [I] gave.” $\endgroup$
    – Oiler
    Jul 12 '21 at 22:45
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Standard general method: Write the presentation in abelianized form in a matrix (rows for relators, columns correspond to generators): $$ \left(\begin{array}{rrr}% 2&0&0\\% 0&3&0\\% 4&4&4\\% \end{array}\right). $$ Calculate the Smith Normal form (also called elementary divisor normal form): $$ \left(\begin{array}{rrr}% 1&0&0\\% 0&2&0\\% 0&0&12\\% \end{array}\right)% $$ Thus the structure of $G/[G,G]$ is $C_2\times C_{12}$ (which is the same as the other andwer, using that $12=3\cdot 4$.

Justification for why Smith normal form works: Row transformations correspond to relator changes (replace relator by product with other), column operations to generator changes.

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The presentation of $G/[G,G]$ is achieved by adding the relations $$xyx^{-1}y^{-1}=1,xzx^{-1}z^{-1}=1,yzy^{-1}z^{-1}=1.$$ The commutators $xyx^{-1}y^{-1},xzx^{-1}z^{-1},yzy^{-1}z^{-1}$ and their inverses are the generators of the normal subgroup $[G,G]$. Then $$G/[G,G]\cong\langle x,y,z\ : x^2,y^3,(xyz)^4, xyx^{-1}y^{-1},xzx^{-1}z^{-1},yzy^{-1}z^{-1}\rangle.$$

This is also because the natural homomorphism $ \langle x,y,z\ : x^2,y^3,(xyz)^4\rangle$ to $\langle x,y,z\ : x^2,y^3,(xyz)^4, xyx^{-1}y^{-1},xzx^{-1}z^{-1},yzy^{-1}z^{-1}\rangle$ is surjective and has kernel the normal closure of $[G,G]$, which is $[G,G]$ itself.

The "abelian" presentation would be $$\langle x,y,z:2x=0,3y=0, 4x+4y+4z=0,\\ x+y-x-y=0,x+z-x-z=0,y+z-y-z=0\rangle,$$ which can be simplified with $t=x+y+z$ into $$\langle x,y,t\ :\ 2x=0,3y=0, 4t=0\rangle,$$ and which is $\mathbb Z_2\oplus \mathbb Z_3\oplus \mathbb Z_4$

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  • $\begingroup$ the abuse of language is the problem Mr. downvoter? here the abuse is that $x,y,z$ should be cosets modulo the $[G,G]$. $\endgroup$
    – janmarqz
    Jul 12 '21 at 18:00
  • $\begingroup$ I did not downvote, but this is not the answer to my question. I am more concerned with why this is the abelian presentation. As I stated in my question, I have no trouble finishing off the problem once I justify why that is the abelian presentation. $\endgroup$
    – Oiler
    Jul 12 '21 at 18:24
  • $\begingroup$ @Oiler, thanks for the clarification, I added a few more details, hoping this might help you $\endgroup$
    – janmarqz
    Jul 12 '21 at 19:18
  • $\begingroup$ "The normal closure of $[G,G]$"... Since $[G,G]$ is verbal, it is already normal. The only care is to note that it is generated by commutators, and may include elements that are not themselves commutators, but products thereof. $\endgroup$ Jul 12 '21 at 19:22
  • $\begingroup$ Agree, the elements of $[G,G]$ are words of the form $c_1c_2\cdots c_r$ where each $c_i$ are commutators themselves. $\endgroup$
    – janmarqz
    Jul 12 '21 at 19:26

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