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



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.
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.



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.