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?
Best Answer
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 $|G|=\alpha\beta\gcd(\alpha,\beta,\gamma)$