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In this question How to derive the cosets of $A_4$?

I derived that the left cosets are $$\{ 1\langle (1, 2, 3) \rangle, (143)\langle (1, 2, 3) \rangle, (142)\langle (1, 2, 3) \rangle, (341)\langle (1, 2, 3) \rangle \}$$ I derived the right cosets $$\{ \langle (1, 2, 3) \rangle 1, \langle (1, 2, 3) \rangle (341), \langle (1, 2, 3) \rangle (241), \langle (1, 2, 3) \rangle (143) \}$$ by simply reversing $$a = 1, (143), (142), (341)$$ to get $$b = 1, (341), (241), (143)$$

I think this has to do with properties of conjugation, cosets or subgroups or normal subgroups, but I don't know which. I think $\langle (1, 2, 3) \rangle$ is not normal because the cosets don't correspond.


Remark based on Nicky Hekster's answer: I didn't realize it sooner! $(1, 2, 3)$ is simply $x$ in $S_3$ whose inverse is $x^2$ in both $S_3$ and in $S_4$ (and in $S_n, n\ge 3$ or even $n = 1,2$ if you want to be vacuous).

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This is because if $H \leq G$, $|G:H|=n$, then a set $\{g_1, g_2, \cdots, g_n\}$ represents the left cosets of $H$ if and only if $\{g_1^{-1}, g_2^{-1}, \cdots, g_n^{-1}\}$ represents the right cosets of $H$. This is easy to prove. And remember, writing a cycle from right to left gives you the inverse of this same cycle.

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  • $\begingroup$ Thank you Nicky Hekster! $\endgroup$
    – user198044
    Oct 12, 2018 at 10:34
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    $\begingroup$ You are welcome!! $\endgroup$ Oct 12, 2018 at 11:42

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