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I'm studying groups diving into the fantastic Artin's Algebra book. Actually, he's presenting the S3 group using the matrix representation of 2 permutations:

$x = (1 3 2)$ and $y = (1 2)$

Actually, he correctly claims the whole group can be generated by these ones, like this (prop. 1.17):

$\{1,x,x^2,y,xy,x^2y\}$

To get the above items, we can use the following rules between generators (prop. 1.18):

$x^3=1, y^2=1, yx = x^2y$

From the text, the author suggests these rules can be verified directly. The 1st and 2nd ones are straightforward, but I was trying to have a small demo about the 3rd one, i.e. $yx = x^2y$. Actually, I guess the author means directly suggesting to replace directly the x and y by the related permutations, and see the lhs and rhs actually match as expected. However, I was struggling trying to demonstrate that rule using other axioms. The most I got, is a different way of expressing the rhs, i.e.:

$x^3 = 1 \implies x^2 = x^{-1} \implies x^2y = x^{-1}y$

After a bit of time, I realized this is probably not possible, since those 3 rules must be taken together as axioms as a whole, without getting one of the other two ones.

In the same paragraph, Artin is claiming any possible pattern of xy can be finally expressed as $x^ny^m$, but in this case the original $x^2y$ is already in that form.

Am I correct in my analysis? I mean: is there a way, using pattern reduction by the rule 1) and 2), to desume the third rule $x^2y$ without direct permutation replacement?

Thanks for the help in advance

riccardoventrella
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1 Answers1

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No, there is no way to deduce the third rule from the two first.

The reason for this is that there is a group generated by two element $x$ and $y$ satisfying the first two rules, but which does not satisfy the third. In fact, there is an infinite group like this (while you have seen that if it also satisfies the third rule then it becomes finite).

Tobias Kildetoft
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  • Thanks for your confirmation. Actually I had the same idea bearing in mind, i.e. there are many (actually infinite, as you mentioned) groups satisfying the first 2 rules. So, the third rule is the one which actually nails down the finite character of this group. Highly helpful, thanks. – riccardoventrella Jun 21 '18 at 07:41
  • I was trying to figure out another group having the first two above mentioned rules as presentation, but having a third one actually different from the S3 one I just mentioned above. May suggest one, please, eventually? Thanks for your help. – riccardoventrella Jun 21 '18 at 21:07