I know $H$ is a subgroup. That is easy to verify. But is it a cyclic group? How would I write out the left cosets?

P.S. I used the cycle notation for permutations.


1 Answer 1


No, $H$ is not cyclic since it contains two distinct elements of order $2$, namely $(12)$ and $(34)$. It is an abelian subgroup, and you can easily argue it is isomorphic to the direct product $C_2\times C_2$; the Klein four group. To compute the cosets, you can do as follows. Pick $\sigma \in S_4$ that is not in $H$. Compute $\sigma H$. Then pick $\tau$ that is not in this last coset, and compute $\tau H$, then pick $\eta$ that is not in $\tau U\cup \sigma H$, and so on. You need to end up with $24/4=6$ cosets.

  • $\begingroup$ So if I pick, $(1,2,4,3)$ in group, then $(1,2,4,3)H={(1,2,4,3)(),(1,2,4,3)(1,2),(1,2,4,3)(3,4),(1,2,4,3)(1,2)(3,4)}$ Correct? But isn't this only 4 cosets? $\endgroup$
    – atherton
    Commented Mar 3, 2014 at 6:20

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