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I know that I have $(\sigma(1)\sigma(2))(\sigma(3)\sigma(4)\sigma(5))=(12)(345)$. However, I've only found eight that commute with this permutation, how do I know if I've found them all?

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  • $\begingroup$ Hint: the number elements is equal 7! / (the number of elements with the same cycle structure). $\endgroup$ – Steve D Nov 6 '17 at 3:59
  • $\begingroup$ You haven't${}$. $\endgroup$ – Angina Seng Nov 6 '17 at 4:12
  • $\begingroup$ Not sure what your notation $(\sigma(1)\sigma(2))(\sigma(3)\sigma(4)\sigma(5))=(12)(345)$ means unless you mean that $\sigma \in S_7$. Note that $H$, the subgroup generated by $(12)(345)$ is a cyclic group of order $6$. And the set of all permutations in $S_7$ that commute with $(12)(345)$ form another subgroup $K$ with $H \le K \le S_7 $. This should give you an idea of what $|K|$ might be. $\endgroup$ – Stephen Meskin Nov 6 '17 at 5:14
  • $\begingroup$ See this question, and several more of this type. $\endgroup$ – Dietrich Burde Nov 6 '17 at 20:22
  • $\begingroup$ @SteveD So would it be 24 elements? $\endgroup$ – NoMayoPlz Nov 6 '17 at 22:59

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