# Manipulating Group Presentations: Are my arguments valid?

I'd like to show that the group given by presentation $G=\langle a,b \ | \ a^2b^2 \rangle$ is not isomorphic to any of $\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}$, or $\mathbb{Z} \times \mathbb{Z}$.

For $G \ncong \mathbb{Z}$, I note that $G$ is not free because it has a relator in its presentation.

For $G \ncong \mathbb{Z}/2\mathbb{Z}$, I claim that $G$ cannot be of order 2, since the subgroup $\langle a \rangle$ in $G$ is infinite: if I write $a^n=1$ for some $n \in \mathbb{Z}\ \backslash\{0\}$ then $a^n$ has to be obtained by a finite number of composition of self-inversions, or self-multiplications, or conjugations on the relator $a^2b^2$. All of these operations preserve the presence of $b$, so we can never get to $a^n$.

For $G \ncong \mathbb{Z} \times \mathbb{Z}$ I argue that $G$ is not abelian. The abelianization of $G$ is $G_{ab}=\langle a,b \ | \ a^2b^2, aba^{-1}b^{-1} \rangle$. How do I show $G_{ab}$ is not trivial?

I'm new to presentations and am completely unsure if the above reasonings are allowed or not. Any help appreciated.

• I forgot to say: the first argument is not correct, as $\langle x,y \mid y \rangle \cong \mathbb{Z}$. The second argument can be made to work, though the level of rigor you might want may make writing a solution unpleasant. Generally such argument are difficult. This is the starting point of combinatorial group theory. Feb 22 '18 at 17:37
• If you're going to show that $G$ is not abelian (as you want to do in the third case), then you will deal with all three cases at once. Feb 22 '18 at 21:52

## 2 Answers

One of the most useful things about group presentations is that it's easy to define homomorphisms from the presented group to groups you already understand. Suppose $G$ is generated by $X$ and $R$ is a set of defining relations. Formally this means you have a homomorphism $F(X) \to G$ with kernel the smallest normal subgroup of the free group $F(X)$ that contains $R \subset F(X)$. If you already know a group $H$, you can try to define a homomorphism $G \to H$ by choosing a target for each $x \in X$, say $\phi(x)$. This defines a homomorphism $F(X)$ by mapping a word over $X$ to the word obtained by replacing each occurrence of the letter $x$ by the group element $\phi(x) \in H$. (That this is a homomorphism of $F(X)$ is sometimes the definition of the free group in terms of a universal mapping property.) Whether or not $\phi$ induces a homomorphism $G \to H$ depends on whether or not every relation $r \in R$ is mapped to the identity in $H$.

In any case, an example is probably best. Define a homomorphism from your group to the integers (written additively, though this can be confusing!) by mapping each of $a$ and $b$ to 1. This does not define a homomorphism because the relator $a^2b^2$ is mapped to $2 + 2$ which is not the identity (zero) in the target group. On the other hand, you can find a homomorphism onto the integers by making a different choice. (This will prove your group is infinite.)

As for the abelianization, you simply rewrite your presentation using additive notation and simplify (but don't divide by integers other than $\pm 1$ when you simplify). So, for your group you get the abelian group with generators $a$ and $b$ such that $2a + 2b = 0$. A change of variables will help: let $c = a + b$. Then an equivalent (but better) description of the abelianization follows and you'l see what it is. (This computation will rule out all of your candidates for what group this might be.)

To show that $G$ is not abelian, one can also exhibit a nonabelian quotient. For example, consider the symmetric group $S_3$, and let $a'$ and $b'$ be two different involutions of $S_3$. There is a surjective homomorphism $G\to S_3$ mapping $a\to a'$ and $b\to b'$, so $S_3$ is a quotient of $G$ and $G$ is non-abelian.