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

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 $$\Bbb Z_3$$

• "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 $S_3$"). Nov 4 '19 at 19:24

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

• Is there a different notation for the line starting with sigma? I don't understand that line
– cele
Oct 15 '19 at 23:35
• I've removed the $+$. Is that any better? What's the first bit you don't understand, @cele? Oct 15 '19 at 23:38
• Yes, they are elements of $S_3$ and $\Bbb Z_3$, respectively, @cele. In general, for any sets $A$ and $B$, we have $$A\times B=\{(a, b)\mid a\in A\land b\in B\}.$$ Oct 15 '19 at 23:49
• 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 $\Bbb Z_3$. Oct 16 '19 at 0:05
• Also, $H$ is a Cartesian product, like $A\times B$. Oct 16 '19 at 0:07