Prove that $\langle{a,b\,\vert\,aba^{-1}b^{2}}\rangle$ is not isomorphic to $\mathbb{Z}/3\mathbb{Z}$. Could anyone give me a suggestion to solve this problem about group presentations?
Prove that $\langle{a,b\,\vert\,aba^{-1}b^{2}}\rangle$ is not isomorphic to 
$\mathbb{Z}/3\mathbb{Z}$
 A: Consider the homomorphism from the given group to $(\mathbb Z,+)$ given by the assignment $$a\mapsto 1 \text{ and } b\mapsto 0 .$$
This takes $aba^{-1}b^2$ to $1+0-1+0+0=0$ so is well defined and is surjective because for any integer $n$, the element $a^n$ will map to $n$. 
Unfortunately, $\mathbb Z/3\mathbb Z$ cannot surject on to an infinite group..
A: In your group, $b$ and $b^{-2}$ are conjugated. So let us look for a group $G$ with an element conjugated to its $(-2)$th power. A simple one is the symmetric group $S_5$, in which $x=(12345)$ and $x^{-2}=(14253)$ are conjugated (because they have the same type of factorization into disjoint cycles) There is some $y$ in $S_5$ such that $yxy^{-1}=x^{-2}$. 
Consider now the map going from your group to $S_5$ mapping $a$ and $b$ to $y$ and $x$, respectively. The image has at least order $5$.
A: Let $G$ be the group.  The abelianization $G/[G,G]$ of $G$ has the presentation
$$\langle a,b\mid aba^{-1}b^{2},aba^{-1}b^{-1}\rangle$$
which implies that $b^2=b^{-1}$, so $b^3=1$. It follows that the abelianization is isomorphic to $\mathbb{Z}\oplus \mathbb{Z}/3\mathbb{Z}$, with $a$ generating the first summand and $b$ generating the second.  Therefore, $G$ is not $\mathbb{Z}/3\mathbb{Z}$ (this is an abelian group already, and its abelianization is definitely not $\mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z}$).
A: Define a homomorphism $f : G \to \mathbb Z/6\mathbb Z$ by $f(a) = 1$ and $f(b) = 2$. We have $f(aba^{-1}b^2) = 1 + 2 -1 + 4 \mod 6 = 0$, so it really is a homomorphism. But $f(a)$ has order 6. Since the image of any element of $\mathbb Z/3\mathbb Z$ has order $\leq 3$, $G$ cannot be isomorphic to $\mathbb Z / 3\mathbb Z$.
