Proving Isomorphic to S3

If you have a group with two elements, $x$ and $y$, which satisfy $x^3 = y^2 = 1$, can you conclude that this group is isomorphic to $S_3$? In general, what are some techniques that you can use to prove something is isomorphic to $S_3$ or $S_n$ (without resorting to Cayley tables)?

• No, you cannot even conclude that the group is finite, or that it is non-abelian. – Arthur Aug 22 '17 at 16:07
• See my "Groups whose elements are of order two or three”, American Mathematical Monthly, 79, 1972, 1007-1010 – Ethan Bolker Aug 22 '17 at 16:14

You need to say more about the group to define it. The abelian group of order $6$ satisfies the relations you have, and the infinite group of complex numbers of modulus $1$ under multiplication has elements of order 2 and order 3. In general determining what the group is from its relations is a hard task.

In general, to get to defining a group more closely you would need to say that $x$ and $y$ generate the group. And you would also need to say (currently undefined) how they interact with each other, which would pin the group down.

You cannot conclude. For instance, the product $\mathbb{Z}_2 \times \mathbb{Z}_3 \simeq \mathbb{Z}_6$ satisfies your condition.

There are also infinite groups satisfying it, for instance the free product $$\mathbb{Z}_2 \ast \mathbb{Z}_3 = \langle x, \, y \; | \; x^2=y^3=1 \rangle.$$

In fact, by the universal property of the free product, any group generated by two elements $x$, $y$ satisfying your condition is a quotient of $G=\mathbb{Z}_2 \ast \mathbb{Z}_3$. For example, $$\mathbb{Z}_2 \times \mathbb{Z}_3 \simeq G/N_1,\quad S_3 \simeq G/N_2,$$ where $N_1$ is the normal closure of the subgroup generated by $xyx^{-1}y^{-1}$ and $N_2$ is the normal closure of the subgroup generated by $xyx^{-1}y$.

If you impose the condition $xy = yx^2$, then you can conclude that. Indeed the elements of this group can all be written has $y^m x^n$, $m = 0,1$, $n=0,1,2$. Then its order is $\le 6$. If it is abelian, then $yx^2 = x^2y$ so $xy = x^2y$ and $x = 1$, which is impossible. The only non-abelian group of order $\le 6$ is $S_3$.

• <code> If you were given the condition that the group containing $x,y$ has order 6, could you also make the conclusion? <code> Thanks! – mathtm Aug 22 '17 at 16:22
• @MessyTiger no, you could have commutativity, in which case your group is $\mathbb Z_2 \oplus \mathbb Z_3$. – Cauchy Aug 22 '17 at 16:24