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I'm having a bit of trouble understanding group presentations. For example, I'm reliably informed that the group

$$ \langle x, y \mid x^2=y^3 \rangle $$

is not the trivial group, but I don't see why not? Why couldn't it be?

Any help appreciated, thanks!

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Hint: In the free group on $x$ and $y$, you cannot get $x$ in the normal closure of $\langle x^2y^{-3}\rangle$ since $x$ will always have an even power in total in that. – Tobias Kildetoft Feb 3 '14 at 11:47
Would $x$ and $y$ commute? If not, then you have an infinite number of words on the form $xyxy\cdots xyxyx$. In any case, you can't have the trivial group, since the empty word, the word $y$ and the word $yy$ are three different elements. – Arthur Feb 3 '14 at 11:47
Or look up free products with amalgamation. Or, think about the infinite tree with vertices of valancy two and three, but no two vertices of the same valency are adjacent. Your group acts non-trivially on this tree in a natural way, so cannot be trivial. (Or just map onto the cyclic group of order 3 in the natural way, but the other two things i mention tell you about the group in a better way). – user1729 Feb 3 '14 at 11:55
This group in fact seems to be a rather fatty one: the free product $C_2*C_3\;$ (or amalgamation over the trivial subgroup) – DonAntonio Feb 3 '14 at 12:07
OK, I am understanding by $\;x^2=y^3\;$ in fact $\; x^2=1=y^3\;$...this seems to be a matter of notation in groups' presentations, perhaps. – DonAntonio Feb 3 '14 at 12:13
up vote 7 down vote accepted

More generally, any group $G$ defined by a finite presentation with more generators than relations is infinite - in fact $G/[G,G]$ is infinite. That follows from the proof of the fundamental theorem of abelian groups.

You can prove it directly, by showing that there is a nontrivial epimorphism $\phi$ onto ${\mathbb Z}$. Let $\phi:G \to {\mathbb Z}$ be any homorphism, which maps generator $x_i$ to $t_i \in {\mathbb Z}$. Then the conditions $\phi(r)=1$ for the group relations $r=1$, reduce to a system of homogeneous linear equations over ${\mathbb Z}$. If there are more generators than relations, then you have more variables than equations, and so there is always a nontrivial solution over ${\mathbb Z}$, which defines a nontrivial $\phi$.

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Let $z = x^2 = y^3$. This element clearly commutes with $x$ and $y$. Therefore $z$ lies in the center and $\langle z \rangle$ is a normal subgroup. We have: $$ G / \langle z \rangle = \langle x, y \mid x^2 = y^3 = 1\rangle. $$

Now the abelianization of this quotient group is $\Bbb Z_2 \times \Bbb Z_3$, which is clearly non-trivial.

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Very nice indeed +1 – DonAntonio Feb 3 '14 at 12:23

If this was the trivial group, then it would follow that for any group $G$ and $x,y\in G$ with $x^2=y^3$ it follows that $x=y=1$. However that is not true, for example consider elements of order $2$ and $3$ in the symmetric group on $3$ elements. (This example also shows that $x$ and $y$ need not commute, so they do not commute in your group.)

Similarly, since $2\cdot 3=3\cdot 2$, there is a homomorphism from your group to $\mathbb Z$, which sends $x$ to $3$ and $y$ to $2$. This shows that $x$ and $y$ are not of finite order.

(Btw, this is the group of a knot, the $(3,2)$-torus knot, and hence its abelianization is $\mathbb Z$.)

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Another example that comes to mind is $C_2\times C_3$. – Pedro Tamaroff Feb 3 '14 at 11:53
@PedroTamaroff, yes, or even $C_2$. – Carsten S Feb 3 '14 at 12:02

The other two answers talk about homomorphisms. that is, the group has an image which is non-trivial and so the group itself is non-trivial. Another approach is to find something which your group acts on. This is what I will do here. I claim that this tells you a lot more about the group than simply investigating the homomorphisms, although the only evidence I will back this claim up with is that the field of geometric group theory exists!

Firstly, your group is going to act on a tree. It is called a free product with amalgamation, and all such groups act on trees in a non-trivial way. However, the corresponding Bass-Serre tree is not very nice, so I will tell you about a nicer tree upon which the group acts.

Consider the infinite tree where every vertex has degree two or three, and no two vertices of the same degree are incident. Equivalently, take the infinite tree where every vertex has degree three and then split every edge in the middle into two edges with a common vertex. Your group acts on this tree, as if you fix a vertex $V_x$ of valency $2$ and another $V_y$ of valency $3$ then $x$ acts by rotating the tree around $V_x$ while $y$ acts by rotating the tree around $V_y$. As $x^2$ and $y^3$ fix the tree, the group acts on the tree. As this action is non-trivial the group is non-trivial.

EDIT: I should say that this is actually proving that the group has the free product $C_2\ast C_3$ as a homomorphic image, and the action investigates this group. However, the idea is a robust one, and can be used to investigate the whole group. But that is better done in the confines of your own room rather than on the internet - the tree is rather hideously infinite...

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Acting on a set $X$ is exactly the same as mapping in $\mathfrak{S}_X$, so your first paragraph is a bit outlandish. – Najib Idrissi Feb 3 '14 at 12:15
So you are talking about a homomorphism to the automorphism group of a tree, right? :) – Carsten S Feb 3 '14 at 12:15
@nik Yes, you are right - I've added a comment. The point was not that this is not a homomorphism, but rather this tells you more about the group than merely a homomorphism. A homomorphism is not the working part, but a by-product. Also, the method scales up so that you have a homomorphism with trivial kernel. – user1729 Feb 3 '14 at 12:18

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