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The article "Classification of Groups of Order $n \le 8$" proves that there are only 5 different groups of order 8 under isomorphism. In this proof (especially in pages 3--4), it checks the multiplication $ba$ over and over again. I can follow the proof. However, I don't quite understand why $ba$ is so important and sufficient for determining a group (or its multiplication table) in this proof.

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  • $\begingroup$ It is not always sufficient. However, for groups with only a few elements, such arguments work. For $n=8$ the relation $ba=ab^3$ in two generators $a,b$ is very important because of the dihedral group $D_4$ and the quaternion group $Q_8$. $\endgroup$ – Dietrich Burde Mar 14 '17 at 14:09
  • $\begingroup$ @DietrichBurde Yes, you are right. However, I mean why is it sufficient in that particular proof. Sorry, I did not make it clear. $\endgroup$ – hengxin Mar 14 '17 at 14:13
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    $\begingroup$ Well, yes, this is the problem with the question. Studying $ba$ is not enough, but the arguments there are enough, to derive only two possibilities, namely the presentations $\langle a,b \mid a^4=b^4=e,a^2=b^2,ba=ab^3\rangle$, and $\langle a,b\mid a^2=b^4=4, ba=ab^3\rangle$ in the non-abelian case. $\endgroup$ – Dietrich Burde Mar 14 '17 at 14:19
  • $\begingroup$ The best answer one can give, I think, is that you should try to follow the whole proof again, until you see all arguments. It is not just "checking the multiplication $ba$ over and over again".. $\endgroup$ – Dietrich Burde Mar 14 '17 at 14:26
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So let's retrace a piece of the proof, namely when $a\in G$ is of order $4$. You take $b\in G$ which is not in $H:=<a>$ so $H\neq Hb$. And since cosets are disjoint then

$$G=\{e, a, a^2, a^3, b, ab, a^2b, a^3b\}$$

So this presentation already tells you the following:

  • You know how to multiply $a^k$ and $a^m$, the product is $a^{k+m}$
  • You know how to multiply $a^k$ and $a^mb$, the product is $a^{k+m}b$

In other words you have just filled a piece of the multiplication table. All that you need to know is how to multiply $a^mb$ by $a^k$ and how to multiply $a^mb$ by $a^kb$ (reversed order of operands).

But note that if you can express $ba$ as an element of the form $a^k$ or $a^kb$ (just like the author does) then you will be able to do appropriate reductions for almost all (I will get back to one special case) products you are looking for. Note that you always can express $ba$ like this, however not every possibility is valid.

Now $ba$ might not determine the group uniquely. The problem is that you still need to know what $b^2$ is (the only unknown product that does not involve neither $ab$ nor $ba$). But if you do know that then you know everything you need.

For example, if $ba=ab$ then

$$(a^2b)(a^3)=a^2baa^2=a^2aba^2=a^3ba^2=a^4ba=a^5b=ab$$

since $a$ is of order $4$. So you now know what element of $G$ is $(a^2b)a^3$. Similarly entire multiplication table can be filled simply by knowing what $ba$ is.

Another example, let $ba=a^2b$. Then

$$(a^2b)(a)=a^2ba=a^2a^2b=a^4b=b$$

Some combinations might be harder to calculate (e.g. $abab$), you need to assume something additional about $b$ (like for example the author considers the case when $b^2=a^2$).

So all in all $ba$ and $b^2$ uniquely determine $G$ in case when $a$ is of order $4$ and $b\not\in<a>$.

I know that my explanation is not formally perfect but I hope you get the intuition.

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  • $\begingroup$ Here's why can we reduce to cases where there exists an element $a$ of order $4$ (in a group of order $8$). The abelian groups of order $8$ are easily named because of the Fund. Thm. of Finite Abelian Groups. The order of an element divides the order of a finite group, so note (1) if there exists an element of order $8$ the group is cyclic (thus abelian), and (2) if all elements are of order $2$ then the group is abelian. Thus any non-abelian group of order $8$ contains an element of order $4$. $\endgroup$ – hardmath Mar 14 '17 at 15:22
  • $\begingroup$ @hardmath Yes. I'm only referring to the case when $a$ is of order $4$ because that's where the confusing $ba$ comes from. The linked paper contains the proof in other cases as well. $\endgroup$ – freakish Mar 14 '17 at 15:24
  • $\begingroup$ I just left the Comment to benefit future Readers; no incentive to edit your Answer was intended! $\endgroup$ – hardmath Mar 14 '17 at 15:26

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