2

Let $G$ be a group with presentation $$G = \langle a,b,c|a^{\alpha} = b^{\beta}= c^{\gamma} = 1, [a,b] = c, [a,c] = [b,c] = 1 \rangle$$ Where $\alpha$, $\beta$ and $\gamma$ are natural numbers. I want to show that $G$ is finite and compute its cardinal.

First, it's easy to show that $G = \{a^{r}b^{s}c^{t} |r,s,t \in \mathbb{Z}\}$, obviously with non-uniqueness of $r,s$ and $t$. So then one tries to define a map (not necessarily a morphism, we are only interested in cardinality) $$ \begin{array}{rcl} \varphi \colon \mathbb{Z}/\alpha\mathbb{Z}\times\mathbb{Z}/\beta\mathbb{Z}\times\mathbb{Z}/\gamma\mathbb{Z} & \longrightarrow & G\\ (r + \alpha\mathbb{Z}, s + \beta\mathbb{Z}, t + \gamma\mathbb{Z}) & \longmapsto & a^{r}b^{s}c^{t} \end{array} $$ This map is well-defined (thanks to the trivial commutators, the equation $a^{r}b^{s}c^{t} = a^{r^{'}}b^{s^{'}}c^{t^{'}}$ is equivalent to $a^{r-r^{'}}b^{s-s^{'}}c^{t-t^{'}} = 1$) and onto. The problem is the injectivity, and I suspect that to achieve this I'll need to change $\gamma$ by the order of $c$ in $G$. Any ideas?

user_12345
  • 388

1 Answers1

1

I'll give a slightly different approach that I find more intuitive:

(Thanks Derek for pointing out a previous error)

We first use the relations to simplify the presentation:

$a^b=ac$ and $a^c=a$ implies $a=a^{b^\beta}=ac^\beta$ so $o(c)\mid\beta$.

Similarly $(b^{-1})^a=cb^{-1}$ and $b^c=b$ implies $o(c)|\alpha$.

Together with $c^\gamma=1$ this gives $o(c)|\gcd(\alpha,\beta,\gamma)=\delta$

Hence $G=\langle a,b,c|a^\alpha=b^\beta=c^\delta=1,[a,b]=c,[a,c]=[b,c]=1\rangle$.

But this is the presentation for $(C_\alpha\times C_\delta)\rtimes C_\beta$.

Hence $|G|=\alpha\beta\gcd(\alpha,\beta,\gamma)$

Robert Chamberlain
  • 7,202
  • 1
  • 15
  • 24
  • Thank you, clearly answering questions too late, have corrected – Robert Chamberlain Oct 16 '19 at 18:47
  • OK I've deleted my comment, which was full of typos. – Derek Holt Oct 16 '19 at 19:22
  • Thanks! Maybe this is too stupid, but I can't see why this new presentation is the presentation for the semidirect product $(C_\alpha\times C_\delta)\rtimes C_\beta$. – user_12345 Oct 16 '19 at 20:18
  • Not stupid at all! See for example here – Robert Chamberlain Oct 16 '19 at 21:28
  • But which morphism are you taking in the semidirect product? And why do you need to make the gcd appear in the presentation? – user_12345 Oct 18 '19 at 19:07
  • In this case we want to identify $a$ with the generator, say $x$, of $C_\alpha$, $c$ with the generator $y$ of $C_\delta$ and $b$ with the generator $z$ of $C_\beta$. With $b^{-1}ab=ac$ we therefore want $z\mapsto(x\mapsto xy)$ (implicitly $y\mapsto y$). As $o(y)\mid o(x)$ we indeed have $x\mapsto xy$ defines an automorphism of $C_\alpha\times C_\delta$. Since $o(y)\mid o(z)$ we have $z\mapsto(x\mapsto xy)$ defines an homomorphism $C_\delta\to {\rm Aut}(C_\alpha\times C_\delta)$. I prove in the answer that $c^\delta=1$, the order is $|G|=\alpha\beta\delta=\alpha\beta\gcd(\alpha,\beta,\gamma)$. – Robert Chamberlain Oct 18 '19 at 20:53