This is Exercise 1.2.21 of Magnus *et al*'s book on combinatorial group theory.

The Question:
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> Show that the group $$G=\langle a, b\mid a^3, b^3, c=b^{-1}a^{-1}ba, ac=ca, bc=cb\rangle$$ has order $27$.
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Thoughts:
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We can rewrite the presentation for $G$ as $$\langle a, b\mid a^3, b^3, [a, [b, a]], [b, [b, a]]\rangle,$$ where $[x, y]=x^{-1}y^{-1}xy$, by letting $c=[b, a]$, multiplying $ac=ca$ on the left by $a^{-1}c^{-1}$, and multiplying $bc=cb$ on the left by $b^{-1}c^{-1}$. In words, $a$ and $b$ commute with the commutator $[b, a]$.

We have, then, that
$$\begin{align}
e&=a^{-1}[b, a]^{-1}a[b, a] \\
&=a^{-1}[a, b]a[b, a] \\
&=\dots
\end{align}$$ but I don't know where I'm going with that.

I'm reminded of Lie algebras. It's been too long since I've studied them however.

Context:
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I'm studying for a PhD in combinatorial group theory, first year. 

I'm familiar with similar questions: there's a general technique of, say, showing that one can write the elements of a group as a word of some form (in the generators, of course), then performing combinatorial arguments on the powers of the generators. I can't seem to do it here.

What kind of answer am I looking for? Well, a rigorous argument for why the group has order $27$ would be great. Combinatorial group theoretic approaches are strongly preferred. 

I reckon I should be able to solve this problem myself, given enough time. I'm still practicing the basics . . .

Please help :)