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I'm trying to teach myself Cosets. Can anyone help with this question:

Let $G$:= $\Sigma_6$ and $x$ = $(123)(456)$ $\in$ $\Sigma_6$. Determine $C_G$($x$) by finding a set $S \subseteq G$ with $C_G$($x$) = $\langle$ $S$ $\rangle$.

Hint: Consider the subgroup $H$ $\leq$ $C_G$($x$) consisting of all elements $h \in$ $G$ with $(123)^h$ = $(123)$ and $(456)^h$ = $(456)$ and describe the right cosets of $H$ in $C_G$($x$).

Any help would be appreciated. Thanks.

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up vote 2 down vote accepted

The subgroup in the hint is $$H = \{h \in S_6 : (1,2,3)^h = (1,2,3) \text{ and } (4,5,6)^h = (4,5,6)\}.$$

This is a 9 element subgroup generated by $(1,2,3)$ and $(4,5,6)$, so it is (isomorphic to) $C_3\times C_3$.

Also, for all $h\in H$,

$$h(1,2,3)(4,5,6)h^{-1} = h(1,2,3)h^{-1}h(4,5,6)h^{-1} = (1,2,3)(4,5,6).$$

Therefore, $H\leq C_G(x)$.

The centralizer of $x$ is an 18 element group isomorphic to $S_3 \times C_3$ and the cosets of $H$ in $C_G(x)$ are $\{H, gH\}$ where

$$g = (1,4)(2,5)(3,6).$$

Therefore, the set $S$ mentioned in the problem statement is

$$S = \{(1,2,3), (4,5,6), (1,4)(2,5)(3,6) \}. $$

This should be enough for you to go on and prove that the centralizer is the 18 element group claimed.

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I have deleted my answer for now. I'll answer this in a bit detail later next evening as I got to go. – user21436 Feb 26 '12 at 14:39

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