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$$G = \langle x, y : x^2 = 1, y^3 = 1, x^{-1}y^{-1}xy = 1 \rangle$$ What are the elements of this group and how do I find them?

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$G = \langle x,y : x^2=y^3=1 \hspace{0.4cm} \text{and} \hspace{0.4cm} x^{-1}y^{-1}xy=1 \rangle.$

From $x^{-1}y^{-1}xy=1$ we have $xy = yx$, so $G$ is abelian. Therefore, the elements of $G$ are given by $x^i y^j$ for $0 \leq i \leq 1$, $0 \leq j \leq 2$.

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    how did you get that last part -- sorry i'm new to this – 2017-01-18
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    No problem. We have $G$ is abelian, so if we have an element $xyyxyx$ (for example) we can group all $x$'s and all $y$'s. So, we try all possible combinations of the form $x^iy^j$ . Since $x^2=1$ and $y^3=1$, then all possibilities are $1$, $x$, $y$, $y^2$, $xy$, $xy^2$. – 2017-01-18
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    how do we know that the two generators commute? – 2017-01-18
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    $x^{-1}y^{-1}xy=1 \Rightarrow y^{-1}xy=x \Rightarrow xy=yx$. – 2017-01-18
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You know there are two generators, one of order two and the other of order 3. Additionally, you know that the two generators commute, so the group is Abelian. Thus if you were to reduce, say, an element like $g = xyxxxyyyxyyyx,$ you would just move the $x$'s and $y$'s together and then reduce the number of $x$'s mod 2 and the number of $y$'s mod 3. So the above $g$ would reduce to $y$ since there are $6$ $x$'s and $7$ $y$'s. Thus you can write any group element uniquely as $x^i y^j$ where $i=0,1$ and $j=0,1,2. $

The $x$ and $y$ don't really talk to each other, and they behave like the cyclic groups $\mathbb{Z}_2$ and $\mathbb{Z}_3.$ This makes the group a direct product $$ G = \mathbb{Z}_2\times \mathbb{Z}_3$$

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    I prefer this answer, because it makes it clear that (e.g.) the group is nontrivial. – 2017-01-18
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    Therefore G = {I, x, y, y2, xy, xy2}? – 2017-01-18
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    @ParthVader Yeah, those are all the elements in terms of the generators. – 2017-01-18
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    how do we know that the generators commute? – 2017-01-18
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    @ParthVader the relation $x^{-1}y^{-1}xy = I$ can be rewritten $xy=yx.$ – 2017-01-18
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    explain algebraically? – 2017-01-18
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    uhh, you mean you want me to walk you through the rearrangement of the formula? Multiply both sides on the left by $x$ and then on the left by $y.$ You get $yxx^{-1}y^{-1}xy = yxI = yx$ and the left hand side collapses to $yxx^{-1}y^{-1}xy =yIy^{-1}xy = y y^{-1}xy = Ixy = xy.$ So $xy=yx$. – 2017-01-18