# Writing $G/[G, G]$ as a direct product of cyclic groups

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?

• 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. Jul 12 '21 at 16:50
• Can you elaborate on what type of "justification as to why" you would be looking for? Jul 12 '21 at 16:58
• @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 $R$. Jul 12 '21 at 18:27
• Your comment does not seem to contain a question, so I am at a loss as to what it is you are looking for. Jul 12 '21 at 18:30
• @ArturoMagidin I had the same thought, but I guess I wanted more details as to why “reduces to the conditions [I] gave.” Jul 12 '21 at 22:45

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.

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 $$\mathbb Z_2\oplus \mathbb Z_3\oplus \mathbb Z_4$$

• the abuse of language is the problem Mr. downvoter? here the abuse is that $x,y,z$ should be cosets modulo the $[G,G]$. Jul 12 '21 at 18:00
• 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. Jul 12 '21 at 18:24
• @Oiler, thanks for the clarification, I added a few more details, hoping this might help you Jul 12 '21 at 19:18
• "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. Jul 12 '21 at 19:22
• Agree, the elements of $[G,G]$ are words of the form $c_1c_2\cdots c_r$ where each $c_i$ are commutators themselves. Jul 12 '21 at 19:26