# 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 $|G|$.

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Do you mean for the relations to read $a^2 = b^3 = e$, where $e$ is the identity? –  Sammy Black Mar 22 '13 at 9:07
Also, the trivial group $G=\{e\}$ with $a=b=e$ satisfies your conditions. –  Stefan Mar 22 '13 at 9:29
If this is homework please tag it as such –  Alexander Gruber Mar 22 '13 at 9:39
you should post the problems carefully –  Mathematician Mar 22 '13 at 9:43
The author's original question seemed to imply that we are free to assume $G$ is a group. We should clarify as the question only makes sense with such an assumption. –  user47805 Mar 22 '13 at 9:59

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 $abab^2ab^2abababa$.

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That free product is known to be isomorphic to the quotient of $SL_2(\mathbb{Z})$ by its center $\langle -I_2\rangle$. Clearly an infinite group. –  Jyrki Lahtonen Mar 22 '13 at 9:50

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.

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I think we have two answer for this question . 1) $|G|=6$ when G is a finite group . 2) G is infinite when it's free group . –  Hung nguyen Mar 22 '13 at 9:59
@Hungnguyen There are many finite groups of order greater than $6$ generated by an element of order $2$ and an element of order $3$. For example $A_4$ is generated by $(12)(34)$ and $(123)$. The only way the question makes sense is if it's a group presentation, and in that case, the group is not finite. –  Alexander Gruber Mar 22 '13 at 20:35
Thanks Alexander Gruber . –  Hung nguyen Mar 29 '13 at 1:33

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)

$(1,2)^2=e$

$(1,2)=(1,2)$

$(1,2,3)=(1,2,3)$

$(1,2)(1,2,3)=(2,3)$

$(1,2,3)(1,2)=(1,3)$

$(1,2,3)^2=(1,3,2)$

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The symmetric group needs an addition relation, such as $ba = ab^2$. –  Sammy Black Mar 22 '13 at 9:38
Yes, but the group satisfies all given conditions. With the edit you are correct. –  user47805 Mar 22 '13 at 9:40
The notation involving the angle brackets usually suggests a group presentation, which is a quotient of the free group on the given set of symbols (the generators) by the normal subgroup generated by expressions on the right (the relations). –  Sammy Black Mar 22 '13 at 9:43
I believe the original question was something along the lines of "If G is a group that satisfies..." in which case the above answer holds. With the edit to the question, you are correct and we require additional assumptions about G. –  user47805 Mar 22 '13 at 9:48
$G=\infty$ because $\forall x\in G$ we has $x=(ab)^n\quad (\fofall n \in \mathbb{Z}$ . With $S_3$ we need add a relation is $(ab)^2=id$ –  Hung nguyen Mar 24 '13 at 2:00