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As an exercise I am trying to prove that the group $$G = \langle a,b,c \mid ac = ba, ab=ca, bc=ab\rangle$$ is infinite and non-abelian. Moreover, the author claims that its center has finite index.

I have tried to find a group which is (a subgroup of) this abstract group, but to no advance. I have seen similar posts on the forum, but their answers were rather specific to the considered group, and did not provide a general approach.

user1008978
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    I don't think you should begin by looking for subgroups, but rather look for quotients. I (no expert) would start by investigating $G/G'$, that is looking to see what happens if $a,b,c$ commute. That would at least show that $G$ was infinite. – ancient mathematician May 17 '24 at 08:44
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    $G$ is infinite, because $\mathbb{Z}$ is its quotient, by adding $a=b=c$ relationship. Showing that it is nonabelian seems harder though. – freakish May 17 '24 at 08:46
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    Try showing that $S_3$ is a quotient. – Sean Eberhard May 17 '24 at 08:57
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    Send $G\mapsto S_3,$ the symmetry group on $3$ elements, with $a\mapsto(12), b\mapsto(13), c\mapsto (23).$ – Thomas Andrews May 17 '24 at 09:08
  • I think that there is no general approach. The word problem for groups is undecidable. Though of course there are simple sufficient conditions for a group being infinite, e.g. if it can be shown that one of the generators has infinite order. – Charles Hudgins May 17 '24 at 22:03
  • In fact the subgroup $\langle a^2\rangle$ is central, and you can show the quotient is $S_3$. – Steve D May 18 '24 at 04:09

2 Answers2

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You've already gotten an answer, but I wanted to discuss the structure of $G$ a bit more. The first relation allows you to write $b=aca^{-1}$, so you can remove the generator $b$ by simply replacing each occurrence with the above word, to get the presentation \begin{equation} G\cong\langle a,c\mid [a^2,c]=1, [a,c^{-1}]=c^{-1}a\rangle \end{equation}

This shows $a^2\in Z(G)$. From the general fact $[a^2,c^{-1}]=[a,c^{-1}]^a[a,c^{-1}]$, in our group this shows (using the above relations) that $a^2=c^{2}$. So \begin{equation} G/\langle a^2\rangle\cong \langle a,c\mid a^2=c^2=1, [a,c^{-1}]=c^{-1}a\rangle \end{equation}

Because $a\equiv a^{-1}$ and $c\equiv c^{-1}$ in this quotient, that last relation is equivalent to $(ac)^3=1$. So $G/\langle a^2\rangle$ is isomorphic to the dihedral group of order $6$, otherwise known as $S_3$. Because $Z(S_3)=\{1\}$, we also know $Z(G)=\langle a^2\rangle$.

We can even give a full description of the structure of $G$. Because $G/G'\cong\mathbb{Z}$ (a free group), we actually have $G\cong G'\rtimes\mathbb{Z}$. Since $[G:Z(G)]$ is finite, we also know $G'$ is finite. Because $G$ is non-Abelian and $G'$ maps to $A_3\le S_3\cong G/Z(G)$, it must be that $G'\cong \mathbb{Z}/3\mathbb{Z}$, so that \begin{equation} G\cong (\mathbb{Z}/3\mathbb{Z})\rtimes \mathbb{Z} \end{equation} with the generator of $\mathbb{Z}$ acting via inversion.

This semidirect product structure can also be seen from the presentation. Let $d=c^{-1}a$. Then the relation $[a,c^{-1}]=c^{-1}a$ can be rewritten as $d^2=ada^{-1}$. The relation $[a^2,c]=1$ can similarly be rewritten as $[a^2,d]=1$. So $G$ has presentation \begin{equation} G\cong\langle a,d\mid [a^2,d]=1, ada^{-1}=d^2\rangle \end{equation}

From these relations we derive \begin{equation} d^4 = ad^2a^{-1} = a^2da^{-2} = d \end{equation} so that $d^3=1$. Our new presentation $\langle a,d\mid d^3=1, ada^{-1}=d^{-1}\rangle$ is exactly our semidirect product.

Steve D
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Its abelianisation is

$$\langle a,b,c\mid a=b=c\rangle^{{\rm ab}}\cong \Bbb Z$$

and $G\twoheadrightarrow S_3$ (see the comments).

Shaun
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