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Compute the left cosets of $\langle (12)\rangle\times\langle[1]\rangle$ in $S_3\times \Bbb Z_3$

Ok, so I know $\langle (12)\rangle$ is in $S_3$ and $\langle[1]\rangle$ is in $\Bbb Z_3$

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  • $\begingroup$ "Ok, so I know $\langle (12)\rangle$ is in $S_3$ and $\langle [1]\rangle$ is in $\Bbb Z_3$." Be careful: Your use of "is in" is not rigorous. It could be interpreted as, say, $\langle (12)\rangle\in S_3$ or $\langle (12)\rangle\subseteq S_3$ instead of $\langle (12)\rangle\le S_3$ (as in "$\langle (12)\rangle$ is a subgroup of $S_3$"). $\endgroup$
    – Shaun
    Nov 4, 2019 at 19:24

1 Answer 1

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Am I multiplying $(12)$ by $[1]$?

No.

Note that $\langle (12)\rangle=\{(1)(2)(3), (12)\}$, whereas $\langle [1]_3\rangle=\{[0]_3, [1]_3, [2]_3\}$. Thus there are six elements in $H=\langle (12)\rangle\times\langle[1]_3\rangle$.

The left cosets of $H$ in your group are of the form

$$(\sigma, [a]_3)H=\{(\tau, [b]_3)\in S_3\times \Bbb Z_3 \mid (\sigma\tau^{-1}, [a-b]_3)\in H\},$$

noting that $(12)^{-1}=(12)$.

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  • $\begingroup$ Is there a different notation for the line starting with sigma? I don't understand that line $\endgroup$
    – cele
    Oct 15, 2019 at 23:35
  • $\begingroup$ I've removed the $+$. Is that any better? What's the first bit you don't understand, @cele? $\endgroup$
    – Shaun
    Oct 15, 2019 at 23:38
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    $\begingroup$ Yes, they are elements of $S_3$ and $\Bbb Z_3$, respectively, @cele. In general, for any sets $A$ and $B$, we have $$A\times B=\{(a, b)\mid a\in A\land b\in B\}.$$ $\endgroup$
    – Shaun
    Oct 15, 2019 at 23:49
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    $\begingroup$ It means, @cele, that $\sigma$ and $\tau$ differ only by an element in $\langle (12)\rangle$ and $[a]_3-_3[b]_3:=[a-b]_3$ means that $[a]_3$ and $[b]_3$ differ only by an element of $\Bbb Z_3$. $\endgroup$
    – Shaun
    Oct 16, 2019 at 0:05
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    $\begingroup$ Also, $H$ is a Cartesian product, like $A\times B$. $\endgroup$
    – Shaun
    Oct 16, 2019 at 0:07

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