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I'm trying to understand more about the presentation of a group. I understand most of the examples that I have seen but I have come across one that I don't understand. Any help would be greatly appreciated.

Let $\langle x_1,x_2 \ | \ x_1^4, \ x_1^2x_2^2, \ x_1x_2x_1x_2^3 \rangle$ be the presentation of a group G expressed in terms of its generators and relators.

I'm not sure how I should be thinking of this group. How can I decide whether it's abelian and what the order of G is?

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    $\begingroup$ NB: Each group has infinitely many presentations, so there is no "the" presentation. $\endgroup$ – Shaun May 18 '20 at 20:01
  • $\begingroup$ In what sense do you not understand? What's the very first thing about it you need help with? $\endgroup$ – Shaun May 18 '20 at 20:03
  • $\begingroup$ Here's one possible way forward: can you find 'an' order of $x_2$ — that is, an $n$ such that $x_2^n=e$? $\endgroup$ – Steven Stadnicki May 18 '20 at 20:33
  • $\begingroup$ I don't want to detract from Steven Stadnicki's stellar answer, but in questions of this kind its traditional for someone to point out: You cannot answer these questions in general. That is, given an arbitrary finite presentation you cannot determine if the group it defines is abelian, or finite, or even trivial (see, for example, here)! However, there are lots of tricks, as Steven's answer explains, and there are one-way procedures which will only terminate if your group is finite, abelian, etc., as Chris Custer's answer suggests. $\endgroup$ – user1729 May 19 '20 at 13:13
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One potentially-useful perspective on group presentations is that they allow us to think of elements in terms of other elements, and to move elements around. For instance, in this case the relation $x^2y^2$ (I'm going to rename $x_1=x, x_2=y$ just to avoid excessive indices floating around) says that $x^2=y^{-2}$ (you can see this by, for instance, multiplying both sides of $x^2y^2=e$ on the right by $y^{-2}$); in particular, we can use this to see that

$$\begin{align} e&=x^4\\ &=x^2x^2\\ &=y^{-2}y^{-2}\\ &=y^{-4}, \end{align}$$ so $y$ has order (at most) $4$ as well. This means that the last relation can be expressed in the form $xyxy^{-1}$. Now, by doing a similar treatment we can write this as $yx=x^{-1}y$. Let's look at what this means; this lets us move an $x$ 'leftwards' past any positive power of $y$. For instance, we could write $y^2xy$ as

$$\begin{align} yyxy& = y(yx)y\\ &=y(x^{-1}y)y\\ & =yx^{-1}y^2. \end{align}$$

But the fact that $x$ and $y$ both have order four means that there are only 'positive' powers of $x$ or $y$; where we put $x^{-1}$ we might as well say $x^3$. In other words, $y^2xy=yx^3y^2$. But now this is just $(yx)x^2y^2$, so we can write it as $x^3yx^2y^2$. And we can keep moving powers of $x$ to the left, getting

$$\begin{align} x^3yx^2y^2&=x^3(yx)xy^2\\ &=x^3(x^3y)xy^2\\ &=x^3x^3(yx)y^2\\ &=x^3x^3(x^3y)y^2\\ &=x^3x^3x^3y^3\\ &=x^9y^3\\ &=xy^3. \end{align}$$

Can you see how to use this to write every element in the form $x^ay^b$ for some (positive) $a$ and $b$? Once you've done that, then (as hinted above) you can find maximum values of $a$ and $b$ for distinct elements; in particular, this will show that the group is finite. From here, I like just computing a product table and seeing if things 'look familiar'.

You may still need to be a little careful of unexpected cross-relations; for instance, $x^3y$ and $xy^3$ are both in 'canonical form', but they're also equal (why?), so there has to be more to the canonical form than just $a$ and $b$ being small enough...

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  • $\begingroup$ @Harry2000 Well, you'd have to first prove that $x\neq x^{-1}$ (that is, $x^2\neq1$). Currently we just have that $x^4=1$, which only implies that the order of $x$ divides $4$ (in particular, it may be that $x^2=1$). One way of proving that $x^2\neq1$ would be to find a concrete group (e.g. a dihedral group or a cyclic group) which satisfies the relations and where $x^2\neq1$. As we have a bound on the order of the group ($|G|\lneq16$), and we know that the order of the group is even, this is do-able. $\endgroup$ – user1729 May 19 '20 at 16:31
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    $\begingroup$ @Harry2000 How do you know that $yxy^{-1}\neq x$? I'll try and expand my answer to explain this, but the final step here is to, as mentioned, map to some concrete group that satisfies the relations. $\endgroup$ – Steven Stadnicki May 19 '20 at 17:26
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There's the Todd-Coxeter algorithm which will enable you to determine the group's order, in case it is finite. Furthermore I trust you will be able to see if it's abelian. See "Todd–Coxeter algorithm - Wikipedia" https://en.m.wikipedia.org/wiki/Todd%E2%80%93Coxeter_algorithm.

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