# Number of elements in an orbit and proving isomorphism.

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?

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

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 $$S_7$$.