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Let $G=\langle x,y,z\mid x^2=y^2z^2\rangle$.

What is the abelianization of this group? (Also, is there a general method to calculate such abelianizations?)

Update: I know how to get a presentation of the abelianization by adding relations like $xy=yx$ and so on. However is it possible to express it as a direct sum of cyclic groups as per the fundamental theorem for finitely generated abelian groups?

Thanks.

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    $\begingroup$ The given group is a quotient of the free group $F_3$ on three generators; consider it how it behaves under the abelianization map $F_3 \to \mathbb{Z}^3$. (Note, for example, that $G^{ab}$ is the maximal abelian quotient of $G$.) $\endgroup$ – anomaly Aug 20 '18 at 2:27
  • $\begingroup$ @anomaly Thanks. I can get a presentation of the abelianization by adding relations like $xy=yx$ and so on. However is it possible to express it as a direct sum of cyclic groups as per the fundamental theorem for finitely generated abelian groups? $\endgroup$ – yoyostein Aug 20 '18 at 2:44
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    $\begingroup$ keywords: Tietze transformations $\endgroup$ – janmarqz Aug 20 '18 at 3:00
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    $\begingroup$ Do you know a proof for the fundamental theorem of finitely generated abelian groups? The one I know is very much of the flavor, "Give me a presentation for the abelian group, and I will decompose it into primitive factors" - maybe your proof does that too? $\endgroup$ – Milo Brandt Aug 20 '18 at 3:12
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    $\begingroup$ @janmarqz I would say that the keywords here are not Tietze transformations, but Smith Normal Form. $\endgroup$ – Derek Holt Aug 20 '18 at 7:08
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You can rewrite your relator such that it has $0$ exponent sum in two of the generators, as the map $x\mapsto xyz, y\mapsto y, z\mapsto z$ is a Nielsen transformation: $$ \begin{align*} \langle x, y, z\mid x^{2}=y^2z^2\rangle &\cong\langle x, y, z\mid x^{2}z^{-2}y^{-2}\rangle\\ &\cong\langle x, y, z\mid (xyz)^{2}z^{-2}y^{-2}\rangle \end{align*} $$ Under the abelinisation map we then get the group: $$ \begin{align*} \langle x, y, z\mid (xyz)^{2}z^{-2}y^{-2}\rangle^{ab}&=\langle x, y, z\mid x^2\rangle^{ab}\\ &\cong \mathbb{Z}^2\times(\mathbb{Z}/2\mathbb{Z}) \end{align*} $$


This is a specific case of a more general phenomenon, where one can adapt the Euclidean algorithm to rewrite using automorphisms a word $W\in F(a, b, \ldots)$ such that it has zero exponent sum in all but one of the relators. For example, writing $\sigma_x$ for the exponent sum of the relator word in the letter $x$: $$ \begin{align*} &\langle a, b\mid a^6b^8\rangle&&\sigma_a=6, \sigma_b=8\\ &\cong\langle a, b\mid (ab^{-1})^6b^8\rangle&&\text{by applying}~a\mapsto ab^{-1}, b\mapsto b\\ &=\langle a, b\mid (ab^{-1})^5ab^7\rangle&&\sigma_a=6, \sigma_b=2\\ &\cong\langle a, b\mid (a(ba^{-3})^{-1})^5a(ba^{-3})^7\rangle&&\text{by applying}~a\mapsto a, b\mapsto ba^{-3}\\ &\cong\langle a, b\mid (a^4b^{-1})^5a(ba^{-3})^7\rangle&&\sigma_a=0, \sigma_b=2 \end{align*} $$ You can think of this as a "non-commutative Smith normal form", but it is more useful in this context than the Smith normal form as it gives you more information than just the abelianisation. For example, it is used in the HNN-extension version of the Magnus hierarchy ($a$ is the stable letter, and the associated subgroups are free by the Freiheitssatz; see J. McCool and P. Schupp, On one relator groups and HNN extensions, Journal of the Australian Mathematical Society, Volume 16, Issue 2, September 1973 , pp. 249-256 doi).

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    $\begingroup$ Seems like overkill not to abelianize first and just work in the category of abelian groups. $\endgroup$ – C Monsour Aug 20 '18 at 13:35
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    $\begingroup$ @CMonsour Probably, but this is what I did when I solved the problem :-) [Also, as I said in the post, the general idea of reducing exponent sums to $0$ has applications beyond abelianisations.] $\endgroup$ – user1729 Aug 20 '18 at 13:39
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I would think of it as starting with $\Bbb{Z}^3=\langle x,y,z\rangle $ and then quotienting out the cyclic subgroup $\langle x^{-2}y^2z^2\rangle$. You can see easily that $x^{-1}yz$, $y$, and $z$ are an alternate set of generators for $G$, so you get $\Bbb{Z}^2\times \Bbb{Z}/2\Bbb{Z}$ generated by $y, z,$ and $x^{-1}yz$ when you take the quotient.

There is no need to know anything about non-abelian groups, since you can abelianize first and then mod out by the relation. (Abelianization is just modding out by some relations, and it doesn't matter which relations you mod out first--you get to the same place in the end.)

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If we have $$\langle x,y,z\ |\ x^{-2}y^2z^2=1\rangle^{\rm ab}=\langle x,y,z\ |\ x^{-2}y^2z^2=[x,y]=[x,z]=[y,z]=1\rangle$$ with one of the Tietze moves $$\langle x,y,z,t\ |\ t=x^{-1}yz, t^2=[x,y]=[x,z]=[y,z]=1\rangle$$ With another, now we arrange $$\langle x,y,z,t\ |\ x=yzt^{-1}, t^2=[x,y]=[x,z]=[y,z]=1\rangle$$ and finally get $$\langle y,z,t\ |\ t^2=[y,z]=1\rangle$$ which clearly is $\Bbb Z+\Bbb Z+\Bbb Z_2$.

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