Finding the order of a group and isomorphism class from the presentation

Question

Let a group be given by $G = \{<a,b> | ab = ba^2, a^3=b^2 = e\}$ where e is the identity. How do I show that the order of G is 6 and determine the isomorphism class.

Answer 1

At the time of writing, there was another answer simply stating that this is the dihedral group on $6$ elements.

Observe that $G$ has at most $6$ elements, since any product of the symbols $a$ and $b$ can be simplified into one of the following six forms by repeatedly applying the rules of the group presentation: $$ e,a,a^2,b,ab,a^2b. $$ On the other hand, there are at least six elements, since $a$ has order $3$ the number of elements is a multiple of $3$, and similarly since $b$ has order $2$ the number of elements is a multiple of $2$. Thus there are exactly $6$ elements.

Finally, observe that the group is non-abelian, since $ab^2$ and $ab$ are distinct means that $ab^2\not=b^2a$. There is only one non-abelian group of order $6$, the symmetric group on $3$ elements.

An explicit bijection is given by sending $a$ to any $3$-cycle and $b$ to any $2$-cycle.

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Regarding the drive-by downvote(s), I assume it is due to confusion regarding the dual answers involving the symmetric group on the one hand, and the dihedral group on the other - which appear (on the surface) to be two different groups. In fact, both answers are correct: the unique non-abelian group of order $6$ is isomorphic to both the symmetric group $S_3$, and the dihedral group with six elements.

Let me explain what is going on here in more detail, since it is actually quite nice. For any natural number $n\in\mathbb N$, we have the following concrete representation of the dihedral group. Take $n$ points equally spaced around the unit circle in the plane. Then the dihedral group of order $2n$ is the group of rigid symmetries of these $n$ points. It consists of rotations (by multiples of $2\pi/n$) and reflections. On the other hand, the symmetric group $S_n$ consists of all ways of permuting these $n$ points - and in particular, includes the rigid symmetries as a special case. When $n=4$ and higher we can find permutations that don't come from rigid symmetries: for instance, the permutation leaving one side of a square fixed and interchanging the corners of the side does not arise from a rigid symmetry, since it does not preserve distances between the pairs of corners. But when $n=3$, a magical thing happens: all permutations arise from rigid symmetries! Thus the symmetric and dihedral groups coincide in the special case we are considering.