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 $27$. This is Exercise 1.2.21 of Magnus et al's book on combinatorial group theory.
The Question:

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 $27$.

Thoughts:
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:
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 :)
 A: Slightly more generally, the last two relations essentially tell you that the group has nilpotency class at most $2$. Nilpotent groups always allow you to collect elements in a nice way, a little bit like  C Monsour  said in the comments. You first collect generators to the left (say) at the cost of creating some commutators. You then collect first-level commutators, at the cost of creating higher commutators. And so on, at some point, the process stops, because of nilpotency.
It also turns out that, restrictions on the order of elements give restrictions on the order of their commutators.
For example, if $c=[a,b]$ as here, then $a^b=ac$. Now, for every integer $n$, we have $(a^n)^b=(a^b)^n=(ac)^n$. If $a$ and $c$ commute, then we get  $(a^n)^b=(ac)^n=a^nc^n$. So, if $a^n=1$, then also $c^n=1$.
(This is very similar to the argument by Nelyudov (user584025), but I think a bit more enlightening. The argument is also more general:  if we have $c=[a,b]$, then  $[a,c]=1$ and $a^n=1$ already imply $c^n=1$. So we don't need to use $b^n=1$ or $[b,c]=1$ for this conclusion.)
A: $$\begin{align}
c&=b^{-1}a^{-1}ba\\
&=b^2a^2ba\\
&=b^2a^2b(c^{-1}c)a\\
&=c^{-1}b^2a^2bca\\
&=c^{-1}b^2a(abc)a\\
&=c^{-1}b^2aba^2\\
&=c^{-2}b^2(abc)a^2\\
&=c^{-2}b^3a^3\\
&=c^{-2}
\end{align}$$
A: This is the presentation for $U(3,3)$, the Heisenberg group over $\Bbb{Z}/3$ of order $3^3=27$.
If we take $a,b\in G$ such that $aZ(G),bZ(G)$ generate $G/Z(G)$, then $1\not= [a,b]$ generates $Z(G)$, so $a,b$ generate $G$.  If $a,b$ both have order $3$, we obtain our presentation $\langle a,b\mid a^3,b^3,[a,b]=[a,b]^a=[a,b]^b\rangle$.  
Reference: K. Conrad's article Groups of order $p^3$.
