A finitely presented group 
Given the presented group 
  $$G=\Bigl\langle a,b\Bigm| a^2=c,\ b(c^2)b,\ ca(b^4)\Bigr\rangle,$$
  determine the structure of the quotient $G/G'$,where G' is the derived subgroup of $G$ (i.e., the commutator subgroup of $G$).

Simple elimination shows $G$ is cyclic (as it's generated by $b$) of order as a divisor of $10$, how to then obtain $G/G'$?  Note $G'$ is the derived group, i.e it's the commutator subgroup of $G$.

Indeed, the group $G/G'$ is generated by $bG'$: let $\alpha$ denote the image of $a$ in $G/G'$ and $\beta$ the image of $b$. Then we have the relations $\alpha^4\beta^2 = \alpha^3\beta^4 = 1$; from there we obtain
$$\beta^2 = \alpha^{-4} = \alpha^{-1}\alpha^{-3} = \alpha^{-1}\beta^{4},$$
so $\alpha = \beta^{2}$. And therefore $\alpha^4\beta^2 = \beta^8\beta^2 = \beta^{10}=1$. So the order of $\beta$ divides $10$. Therefore $G/G'$ is a quotient of $\langle x\mid x^{10}\rangle$, the cyclic group of order $10$.
Now consider the elements $x^2$ and $x$ in $K=\langle x\mid x^{10}\rangle$. We have $x\Bigl( (x^4)^2\Bigr)x=1$ and $x^4x^2(x^4) =1$. Therefore, there is a homomorphism $G\to K$ that maps $a$ to $x^2$ and $b$ to $x^{10}$, which trivially factors through $G/G'$. Therefore, $G/G'$ has the cyclic group of order $10$ as a quotient.
Since $G/G'$ is a quotient of the cyclic group of order $10$ and has the cyclic group of order $10$ as a quotient, it follows that $G/G'$ is cyclic of order $10$ (generated by $bG'$). 
