Proving $G := \langle a, b, c \mid abc^{-1}a^{-1}, bcb \rangle$ is not isomorphic to $H := \langle a, b \rangle$ I'm trying to prove that $G := \langle a, b, c \mid abc^{-1}a^{-1}, bcb \rangle$ is not isomorphic to $H := \langle a, b \rangle$. 
If they are isomorphic, then their abelianizations $G/[G, G] = \langle a, b, c \mid abc^{-1}a^{-1}, bcb, abc^{-1}a^{-1}, bcb, aba^{-1}b^{-1}, bcb^{-1}c^{-1} \rangle$ and $H/[H, H] = \langle a, b \mid aba^{-1}b^{-1} \rangle$ are isomorphic. Writing those groups as $\langle a, b, c \rangle / N$ and $\langle a, b \rangle /M$ for ease of notation, we can write any element in $G/[G, G]$ as $a^nb^mc^kN$ due to the words $aba^{-1}b^{-1}, bcb^{-1}c^{-1}$ and then as $a^nc^{m + k}N$ due to the word $abc^{-1}a^{-1}$. I don't see how we proceed from here.
 A: You are almost done:
$H/[H,H]\cong\mathbb{Z}^2$
In $G/[G,G]$, we have $1=abc^{-1}a^{-1}=bc^{-1}$ so $b=c$. Therefore $1=bcb=b^3$ so $G/[G,G]\cong\mathbb{Z}\times  \mathbb{Z}/3\mathbb{Z}$
A: If $abc^{-1}a^{-1}=e$ then $bc^{-1}=e$, so $b=c$. But then we have that $c^3=e$, hence the groups are not isomorphic (the free group on two letters does not contain a non-trivial torsion sub group).
Remark (as mentioned in the comments below):  The fact that $b=c$ means that that the group $G$ has the presentation $G=<a,b\,|\,b^3>$.  It is clear that $b\neq e$, since, e.g.,  $G$ has $S_3$ as a quotient group. Thus $G$ contains an element of order $3$.  As the Free group on two letters does not contain any non-trivial element of finite order, it is not isomorphic to $G$. 
A: The notation $H := \langle a, b \rangle$ means the subgroup of $G$ generated by the elements $a$ and $b$. It does not mean the free group on two generators (which might be written $F(a, b)$ or $\langle a, b\mid \rangle$ or $\langle a, b\mid - \rangle$ or $\langle a, b\mid \emptyset \rangle$). Therefore, I disagree with the other two answers, and also with the working in the question.
In fact, $G=H=\langle a, b\rangle$. This is because the generator $c$ can be written in terms of the generators $a$ and $b$ (in particular, the relator $bcb$ implies $c=b^{-2}$).

If you do actually mean that $H=F(a, b)$ then one way of seeing this which is quite different from the other two ways is to eliminating the generator $c$ and obtain the presentation:
$$
G = \langle a, b \mid ab^3a^{-1} \rangle
$$
This word is reduced, and hence $G$ is a proper factor group of $F(a, b)$. As $F(a, b)$ is Hopfian*, it follows that $G\not\cong F(a, b)$.
This method extends to other groups. For example, it proves that the group $K=\langle a, b\mid [a^2, b^2]\rangle$ is not free. However, the methods in the other two answers do not work for $K$: it is torsion free, so we cannot look at the elements of finite order, and its abelinisation is $\mathbb{Z\times Z}$ (the same as $F(a, b)$).
*A group $K$ is Hopfian if $K/N\cong K\Rightarrow N=1$. You can find a proof that free groups are Hopfian here, on Math.SE or here, on Groupprops.
