How do I find $\left|\langle a,b\mid a^2=b^3=e\rangle\right|$? Suppose $G$ is a group satisfying $G=\langle a,b\mid a^2=b^3=e\rangle$. Find $|G|$.
 A: Note:  I am assuming that you intended $G = \langle a, b \,|\, a^2 = b^3 = e \rangle$.
This is not a finite group:  it is the free product $\mathbb{Z}_2 * \mathbb{Z}_3$.  It has distinct elements that are words of any length that alternate between $a$ and either $b$ or $b^2$, such as $abab^2ab^2abababa$.
A: To form a group using a presentation you must write $\langle S | R \rangle$ where $S$ is a set of generators and $R$ is a set of relations among those generators.  $3$ is not in the set of generators in what you have written, so this is not a valid group presentation.
If instead of $3$ you meant to write $1$ or $e$ or $\text{id}$, the group formed does not have finite order, because $a^{-1}b^{-1}ab$ is not present in the relators in any form.  Thus we can form an infinite number of elements that look like $ababababab\ldots$ as we have no way of reducing these words.
A: With the edit to the problem, this answer is 
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Assuming we mean $G=\langle a,b|a^2=b^3=e\rangle$.
Consider $S_3$ where $(1,2)^2=(1,2,3)^3=e$ and $|S_3|=6$.
We have (I'm sure there is an elegant way to do this)
$(1,2)^2=e$
$(1,2)=(1,2)$
$(1,2,3)=(1,2,3)$
$(1,2)(1,2,3)=(2,3)$
$(1,2,3)(1,2)=(1,3)$
$(1,2,3)^2=(1,3,2)$
