On page 42 in his book Algebra, Artin begins describing the symmetric group $S_3$. He claims the six elements of the group are $\{1,x,x^2,y,xy,x^2y\}$. I don't understand how he got these elements. How did he get these 6 elements?
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I define S_3 as the set of all permutations on a set of 3 elements. – Samuel Johnson-Noya Apr 26 '20 at 13:57
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You do know that there are $3! = 6$ such permutations, right? (That will confirm the number of elements in the set, for starters) – The Chaz 2.0 Apr 26 '20 at 13:58
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1Yeah, I know that. I'm just confused as to why he is representing each element as a power of x and y. – Samuel Johnson-Noya Apr 26 '20 at 14:00
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Note that ${1, x, x^2, y, xy, x^2y}=D_3$. – Dietrich Burde Apr 26 '20 at 14:07
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Let $x=(123), y=(23)$. That is, $x$ is a $3$ cycle and $y$ a $2$ cycle, or transposition.
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2Well, $x^2=(132)$. So $x^2y=(132)(23)=(13)$. Play around with it a little to get a feel for it. – Apr 26 '20 at 14:03
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The group $S_3$ has $6$ elements: the identity, $(1\ \ 2)$, $(1\ \ 3)$, $(2\ \ 3)$, $(1\ \ 2\ \ 3)$, and $(1\ \ 3\ \ 2)$. Now, just let $x=(1\ \ 2\ \ 3)$ and let $y=(1\ \ 2)$.
José Carlos Santos
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