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Do we know the structure of the centralizer of any element in $S_n$? Or the centralizer of a permutation which has $q$ disjoint $r$-cycles where $r\leq n$. If we know, can anyone give proof or reference? Thanks.

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Conjugating a cycle by any permutation $\sigma$ yields $$\left(i_1 \ldots i_m\right)^\sigma = \left(\sigma(i_1) \ldots \sigma(i_m)\right).$$ You can prove this without too much trouble, but if you get stuck you can find it in Dummit and Foote. So, the centralizer of any cycle is generated by all $\sigma\in S_n$ which cyclically permute $i_1,\ldots, i_m$.

Now consider the case $\tau = (i_1\ldots i_r)(j_1\ldots j_s)$, where the $i's$ and $j's$ are disjoint. If $r\not= s$ then the centralizer of $\tau$ is simply generated by those permutations which cyclically permute the $i$'s and those which cyclically permute the $j$'s. If $r=s$, however, you must also include the permutation $(i_1j_1)(i_2j_2)\cdots (i_rj_r)$ in the centralizer. Of course, any permutation can be written as a product of disjoint cycles, so you can inductively extend this process to find the centralizer of any element of $S_n$.

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  • $\begingroup$ Can you explain "cyclically permute"? Thank you for your response. $\endgroup$ Dec 26, 2012 at 22:09
  • $\begingroup$ Its action on $i_1 \ldots i_m$ should be cyclic, which just means that $\langle (i_1 \ldots i_m)\rangle$ will be included in the centralizer. So for example if you were finding the centralizer of $(1,3,4)$ in $S_5$, it would be generated by $(1,3,4)$ and $(2,5)$. $\endgroup$
    – Alexander Gruber
    Dec 26, 2012 at 22:13

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