# Centralizer of symmetric group

Let, an element of symmetric group $$S_N$$ is given by $$g=(1)^{N_1}(2)^{N_2}....(s)^{N_s}.$$ Here $$N_n$$ denotes the number of cycles of length $$n$$. Its known that the centralizer of this element is given by $$$$C_g = S_{N_1} \times (S_{N_2} \rtimes \mathbb{Z}_2^{N_2} ) \times \dots\times (S_{N_s} \rtimes \mathbb{Z}_s^{N_s} ).\tag{1}$$$$ I have been able to convince myself this formula gives the correct result when $$g$$ is the identity $$(g=(1)^{N_N})$$ and when $$g$$ is given by $$g=(N)^1$$.

However, lets take a simple case: let's try to find the centralizer of $$(1,2)(3,4)$$ in $$S_4$$. The answer is $$C=\{Id, (1, 2)(3, 4), (1, 2), (3, 4), (1, 3)(2, 4), (1, 4)(2, 3), (1, 4, 2, 3), (1, 3, 2, 4)\}$$.

I don't how I can construct this set using definition (1).

Can anyone walk me through the process please? I tried to construct $$\mathbb{Z}_2^2\rtimes S_2$$. This should be isomorphic to $$D(4).$$ Then I wrote down the elements of $$D(4)$$ in cycle notation but that didn't give me the correct answer.

• Please elaborate on your definition of $g$ because, at least to me, it is not clear. It looks like $g={\rm id}$. – Shaun Mar 20 at 19:37
• @Shaun OP already mentioned that $N_n$ is the number of cycles of length $n$. That is $g$ is decomposed into cycles. – Quang Hoang Mar 20 at 19:40
• That part is clear, yes, @QuangHoang; what's not clear to me is what each $(n)$ means. – Shaun Mar 20 at 19:41
• @Shaun from the notation, that $(n)$ represents a generic cycle of length $n$? – Quang Hoang Mar 20 at 19:42
• (n) is a cycle of lenth n. For example (2,3),(1,4) etc are all denoted by (2) – Shov432 Mar 20 at 19:42

There is a standard way to construct the centralizer of a permutation $$g\in S_N$$ and following this construction we are able to find the formula $$(1)$$. The construction for a general $$g$$ is tedius, I try to write the algorithm and I will make the centrilizer for $$(1,2)(3,4)\in S_4$$

Step $$1$$: Compute the cardinality of $$C_g$$.

$$C_g$$ is the stabilizer of $$g$$ by the conjugation action of $$S_N$$. It's easy to compute the cardinality of the orbit (it's just the cardinality of the conjugation class). Then we have: $$|C_g| = \dfrac{|S_N|}{|orb_{S_N}(g)|}$$

In our case $$g=(1,2)(3,4)$$ and we have: $$|orb_{S_4}(g)| = \binom{4}{2}\dfrac{2!}{2}\cdot \binom{2}{2}\dfrac{2!}{2}\cdot \dfrac{1}{2!} = \dfrac{4!}{2\cdot 2}\cdot \dfrac{1}{2!}$$ I write the cardinality of the orbit in this way because the last $$\frac{1}{2!}$$ represent the way you can choose the position of the transpositions that compose $$g$$ (you will see what I mean in Step $$2$$). Then we obtain: $$|C_g| = \dfrac{4!}{\dfrac{4!}{2\cdot 2}\cdot \dfrac{1}{2!}} = 2^2 \cdot 2!$$ Again think that the first part and the second part are distinct: they represent in some sense $$2$$ different part of the centralizer.

If you use this method to calculate the cardinality of centralizer of a generic permutation $$g = (1)^{N_1}\cdots (s)^{N_s}$$ you will find the formula: $$C_g = 1^{N_1}\cdot 2^{N_2}\cdots s^{N_s} \cdot (N_1)!(N_2)!\cdots (N_s)! = 1^{N_1} (2^{N_2}(N_2)!)\cdot \ \cdots \ \cdot (s^{N_s}(N_s)!)$$

Step $$2$$: Discover two important subgrouos $$H,K \subset C_g$$ related to the computed cardinality.

We define the power subgroup $$H$$ of $$C_g$$ as the group generated by the powers of the cycles that forms the permutation $$g$$. It's easy to see that $$H$$ is a subgroup of $$C_g$$ and it's cardinality is $$|H| = 1^{N_1}\cdot 2^{N_2} \cdots s^{N_s}$$ It's also easy to see that $$H\cong \mathbb{Z_2}^{N_2}\times \cdots \times \mathbb{Z_s}^{N_s}$$

In our case $$g=(1,2)(3,4)$$, then $$H = \{e, (1,2), (3,4), (1,2)(3,4)\}\cong \mathbb{Z_2}^{2}$$

Define the permutation subgroup $$K$$ as the set of permutations inside $$C_g$$ that "permutes by conjugation" the cycles of equal length. To understand what i mean I give you an example: let $$\sigma=(1,2,3)(4,5,6)(7,8,9)$$: an element of $$K$$ is (for example) $$\alpha = (1,4)(2,5)(3,6)$$ or $$\beta = (1,7,4)(2,8,5)(3,9,6)$$; infact: $$\begin{gather} \alpha \sigma \alpha^{-1} = \alpha (1,2,3)(4,5,6)(7,8,9)\alpha^{-1} = (4,5,6)(1,2,3)(7,8,9) = \sigma\\ \beta \sigma \beta^{-1} = \beta (1,2,3)(4,5,6)(7,8,9)\beta^{-1} = (7,8,9)(1,2,3)(4,5,6) = \sigma\\ \end{gather}$$ So $$\alpha$$ and $$\beta$$ are element of $$C_g$$ and they permute the cycles of $$\sigma$$ when you acting by conjugation ($$\alpha$$ switches the first and the second cycle, $$\beta$$ move all three cycles). If you understand what is $$K$$ you are able to see that you can obtain by $$K$$ all the configuration of the cycles of the same length (in the example above, if you named $$(1,2,3)=a, (4,5,6)=b, (7,8,9)=c$$ then $$\alpha$$ is the "permutation" $$(a,b)$$ and $$\beta$$ is the "permutation" $$(a,c,b)$$). Finally we obtain $$K\cong S_{N_1}\times S_{N_2}\times \cdots \times S_{N_s}$$ where $$S_{N_i}$$ is isomorphic to the group that permutes che $$N_i$$-cycles of length $$i$$. Observe that $$|K|=(N_1)!\cdots (N_s)!$$

In our case $$g=(1,2)(3,4)$$, then $$K = \{e, (1,3)(2,4)\}\cong S_2$$.

Step $$3$$: $$H\cap K = \{e\}$$ and $$H$$ is normalized by $$K$$ (i.e. for all $$k\in K$$ $$kHk^{-1} = H$$)

This is the tedius part. If you have a permutation you can do the computation and you can esay prove the two statements: in our case it's obvious that the intersection is only the identity and up to few computation you discover that $$K$$ normalize $$H$$.

I give you just an idea for the general case. Consider $$\sigma = (n)^{N_n}$$ (after some semplification you have to study only this case: you consider the problem restricted on the numbers inside the cycles of length $$n$$). Denote the cycles of $$\sigma$$ as $$a_1,...,a_{N_n}$$ and consider the action of $$C_g$$ over the set of this cycles. You can show that $$H$$ is the kernel of this action and that every element of $$K \backslash \{e\}$$ is not mapped to zero. Then you obtain the two statements of this Step.

Step $$4$$: Conclude that $$C_g \cong H\rtimes K$$

Since $$H\cap K = \{e\}$$, $$|H|\cdot|K| = |C_g|$$ and $$H$$ is normalized by $$K$$ you obtain that $$C_g \cong H\rtimes K$$. After the same semplification mentioned in Step 3 you obtain the formula $$(1)$$.

In our case we have $$C_g \cong H\rtimes K = \mathbb{Z_2}^2\rtimes S_2$$ and if you try to list all the elements you obtain exactly $$C$$ you wrote in your question.