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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)?

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

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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 $xyx^{-1}y$.

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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 $S_3$.

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

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