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I am struggling over an statement from Artin's textbook on Algebra. Its about $S_3$ (the group of all permutations over a set of 3 elements). The elements $1, x$ and $y$ (identity, cyclic permutation and transposition) and the rules $x^3 = 1, y^2=1$ and $yx =x^2y$ are introduced. Then it is stated that $S_3$ contains six elements $\{e,x,x^2;y,xy,yx^2\}$. Now it is stated that one can show by using the cancellation law that all these are distinct. This is what I do not understand. The cancellation law directly gives rise to identities like: iff $ab=ac$ then $b=c$. So how to use these to show there are no identities?

My only idea to solve this was to assume a kind of cancellation inequality-law ($ab \neq ac \Leftrightarrow b \neq c$) and further to assume the inequality of the three (generating) elements ($1,x,y$) and then check if all combinations of elements can be accessed by expanding the three initial rules. That somehow seems to far stretched to be implied by the short statement in the book.

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It should be clear that $x\neq e$ and $y \neq e$, as transpositions do not fix "everything", nor do $3$-cycles (as functions on a $3$-element set).

If $x = x^2$ then by cancellation $e = x$, so $x,x^2$ are distinct. If $x^3 = e$, then if $x^2 = e$ as well, we have $x^2 = x^3$, and by cancellation $e = x$. So we can conclude $e,x,x^2$ are all distinct.

Now $x$ is of order $3$, and $y$ is of order $2$, so they cannot be the same element of $S_3$. Similar reasoning shows $y$ cannot be $x^2$, either.

So $\{e,x,x^2,y\}$ are all different.

It gets easier from here:

$xy = e \implies y = x^{-1} = x^2$, which we ruled out above.

$xy = x \implies y = e$ (but the identity is not a transposition).

$xy = x^2 \implies y = x$ (by cancellation).

$xy = y \implies x = e$ (but the identity is not a $3$-cycle).

So we conclude $\{e,x,x^2,y,xy\}$ are all distinct.

$x^2y = e \implies y = (x^2)^{-1} = x$

$x^2y = x \implies xy = e$ (see above).

$x^2y = x^2 \implies y = e$.

$x^2y = y \implies x^2 = e$.

$x^2y = xy \implies x = e$.

So all of $\{e,x,x^2,y,xy,x^2y\}$ are distinct.

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  • $\begingroup$ OK! And in fact it means going back to the underlying set and working out the inequalities. That was just a bit surprising since reference to the cancellation law seemed to me to suggest that one can already here stick at the more abstract level of group operations. $\endgroup$ – Rudi_Birnbaum Nov 8 '15 at 7:05
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Everything you say is completely spot on, and saying the $6$ elements are all distinct "by the cancellation law" is very poor wording. I think it is intended to mean exactly what you are guessing in the second paragraph: assuming that $1$, $x$, and $y$ are all distinct, you can show that all $6$ elements are distinct. It would be impossible to prove they are all distinct without making some such assumption (or otherwise using information about the group besides just the identities you stated), because if you don't assume $x$ and $y$ are distinct from $1$, then for all you know you could be in the trivial group (note that if $x=y=1$, then the identities $x^3=1$, $y^2=1$, and $yx=x^2y$ are still true).

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    $\begingroup$ Indeed, otherwise the trivial group would satisfy "all the rules". $\endgroup$ – David Wheeler Nov 8 '15 at 6:57
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All this means is that if $y=xy$ for example, then $1=x$ which is a contradiction so $y$ and $xy$ really are different.

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  • $\begingroup$ Does that mean my initial idea is right? $\endgroup$ – Rudi_Birnbaum Nov 8 '15 at 6:53
  • $\begingroup$ @Franky_GTH Yup, you're right. $\endgroup$ – user223391 Nov 8 '15 at 6:54

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