I'm asked to find the number of elements in the orbit of the permutation $\sigma=(123)(45)(67)$ in $S_{7}$, and prove that its centralizer is isomorphic to $\mathbb{Z}_3\times D_4$.

I was able to prove that its orbit contains $420$ elements - all the elements in $G({\sigma})$ (its orbit) are in the form of $(x_{1}x_{2}x_{3})(x_{4}x_5)(x_6x_7)$ so using combinatorics and the fact that for example $(x_{1}x_{2}x_{3})=(x_{3}x_{1}x_{2})=(x_{2}x_{3}x_{1})$ I get that the number of elements is 420 ($\frac{7\times 6\times 5}{3}$ options for $(x_{1}x_{2}x_{3})$, $\frac{5\times 4}{2}$ for $(x_{4}x_5)$ and $\frac{2}{2}$ for $(x_6x_7)$) .

Now I tried finding its centralizer. I was able to find an element with order $4$ - $(6475)$ and an element of order $2$ - $(45)$ and an element of order $3$ - $(123)$. Moreover, I've proven that $\langle(123)\rangle, \langle(6475),(45)\rangle $ are subgroups of $C_{s_{7}}(\sigma)$ (that is the centralizer of $\sigma$ in $S_7$). So if I prove that these are all the elements in $C_{s_{7}}(\sigma)$, I'll get that $\mathbb{Z}_3\times D_4$ is isomorphic to $C_{s_{7}}(\sigma)$. But, using the orbit - stabilizer theorem I get that $|C_{s_{7}}(\sigma)|$=$\frac{|S_7|}{|G(\sigma)|}=\frac{7!}{420}=12\neq|\mathbb{Z_3}\times D_4|=24$.

Where is my mistake?

  • 2
    $\begingroup$ There are different conventions, according to which $|D_4|$ is $4$ or $8$ $\endgroup$ Nov 29, 2020 at 22:28
  • $\begingroup$ @J.W.Tanner according to our professor its magnitude is 8. Are all my calculations correct? $\endgroup$
    – GBA
    Nov 29, 2020 at 22:40
  • $\begingroup$ note that $(x_1x_2x_3)(x_4x_5)(x_6x_7)$ is the same as $(x_1x_2x_3)(x_6x_7)(x_4x_5)$ $\endgroup$ Nov 29, 2020 at 23:05

1 Answer 1


I think your mistake is that there are only $210$ conjugates of $\sigma$ in $S_7$;

you counted each twice, once as $(x_1x_2x_3)(x_4x_5)(x_6x_7)$ and once as $(x_1x_2x_3)(x_6x_7)(x_4x_5)$.

Here is a convenient on-line reference about $S_7$.


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