Confusion about the classification of groups of order $8$

Question

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.

Answer 1

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.