# Why did this reversal from the left cosets of $\langle (1, 2, 3) \rangle$ in $A_4$ give me the right cosets?

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).

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.