MHB Finding the Center of $S_n$: Understanding the Commutativity of Disjoint Cycles

[evinda]

Hello! (Wave)

I want to find the center of $S_n$.

$Z(S_n)=\{ c \in S_n: cg=gc , \forall g \in S_n\}$.Every $c\in S_n$ can be written as a product of transpositions, $$c = \prod_{i = 1}^r (1\, m_i)$$ for some $r$ and some sequence $(m_i)_{1\leqslant i \leqslant r}$, right?

Doesn't this imply that we can consider $g$ to be of the form $(1 \ m)$ for some $m \in \mathbb{N}$?
Because if we want to pick for example $g=(1 2) (2 3)$ then $(2 3)$ can be included at the product of transpositions that represents c, or am I wrong?

Then since $g$ is of the form $(1 m)$ it holds that $g=g^{-1}$.

So $cg=gc \Leftrightarrow c=gcg$.

Let $m \in \mathbb{N}$ such that $g=(1 m)$.

Also let $c=(a_1 a_2 \dots a_n)$.

Is it right so far? Do we have to distinguish now cases for $c$, i.e. if it contains 1 and m, one of them or none of them?

 

[Deveno]

First off, you'll want to make some restrictions on $n$, because $S_2$ is abelian, and $S_3$ is not, so they cannot possibly have even isomorphic centers.

Secondly, I recommend you write an arbitrary permutation as a product of disjoint cycles. Disjoint cycles commute (with each other), so this reduces the problem to finding which cycles can be in the center.

Third, note that a cycle $c \in S_n$ is in the center if and only if $gcg^{-1} = c$ for every $g \in S_n$.

Finally, I believe you may find the following formula useful:

If $c = (a_1\ a_2\ \dots\ a_k)$ and $g \in S_n$ is any permutation, then:

$gcg^{-1} = (g(a_1)\ g(a_2)\ \dots g(a_k))$.

 

[evinda]

First off, you'll want to make some restrictions on $n$, because $S_2$ is abelian, and $S_3$ is not, so they cannot possibly have even isomorphic centers.

How can we show that $S_2$ is abelian but $S_n$ is not for $n\geq 3$ ? (Thinking)

Disjoint cycles commute (with each other), so this reduces the problem to finding which cycles can be in the center.

Why do disjoint cycles commute? Because the order we write them doesn't matter?

Also why does this reduce the problem to finding which cycles can be in the center?

Finally, I believe you may find the following formula useful:

If $c = (a_1\ a_2\ \dots\ a_k)$ and $g \in S_n$ is any permutation, then:

$gcg^{-1} = (g(a_1)\ g(a_2)\ \dots g(a_k))$.

Could you explain to me why the formual holds? (Sweating)

 

[johng1]

Deveno says:

Disjoint cycles commute (with each other), so this reduces the problem to finding which cycles can be in the center.

Perhaps I'm being a bit dense, but I don't see why this is true. It is definitely not true that if commuting elements a and b of an arbitrary group are such that ab is in the center, then each of a and b is in the center.

I would tackle this as follows:
Suppose $z\in Z(S_n)$. Consider the cyclic decomposition of $z$. Suppose there is a cycle of z of length at least 3, say $c=(i,j,k\cdots)$ and $z=cc_1c_2\cdots c_m$ where each $c_i$ is a cycle not containing either of $i$ or $j$. That is $c_i$ commutes with the cycle $(i,j)$. Thus the centralizer of $(i,j)$ contains $c$; the centralizer of $(i,j)$ is a subgroup. But by direct calculation, c does not commute with $(i,j)$. Hence every cycle of z has length at most 2. Now you try and finish the proof.

 

[Deveno]

Deveno says:

Perhaps I'm being a bit dense, but I don't see why this is true. It is definitely not true that if commuting elements a and b of an arbitrary group are such that ab is in the center, then each of a and b is in the center.

I would tackle this as follows:
Suppose $z\in Z(S_n)$. Consider the cyclic decomposition of $z$. Suppose there is a cycle of z of length at least 3, say $c=(i,j,k\cdots)$ and $z=cc_1c_2\cdots c_m$ where each $c_i$ is a cycle not containing either of $i$ or $j$. That is $c_i$ commutes with the cycle $(i,j)$. Thus the centralizer of $(i,j)$ contains $c$; the centralizer of $(i,j)$ is a subgroup. But by direct calculation, c does not commute with $(i,j)$. Hence every cycle of z has length at most 2. Now you try and finish the proof.

That's not quite what I said (although I see where you would think I was implying it).

What I meant (and I apologize for not being clearer) is:

If $g$ commutes with $a$, and it commutes with $b$, it commutes with $ab$.

On the other hand, if $g$ does NOT commute with $a$ (or $b$) it cannot be in the center, so testing whether or not it commutes with $ab$ is a waste of time.

I apologize for the confusion.

 

[evinda]

We have $$\sigma \cdot (1, \ldots, n) \cdot \sigma^{-1} = \big(\sigma(1), \sigma(2), \ldots, \sigma(n)\big) \\ \Rightarrow \sigma \cdot (1, \ldots, n) \cdot \sigma^{-1} \cdot \sigma = \big(\sigma(1), \sigma(2), \ldots, \sigma(n)\big)\cdot \sigma \\ \Rightarrow \sigma \cdot (1, \ldots, n) = \big(\sigma(1), \sigma(2), \ldots, \sigma(n)\big)\cdot \sigma $$

Suppose that $\sigma$ is not the identity.

Then $$\big (\sigma (1), \sigma (2), \dots , \sigma (n)\big )\neq \big (1,2,\dots , n\big )$$

So we have that $$\sigma \cdot (1, \ldots, n) \neq \big (1,\dots , n\big )\cdot \sigma$$

So when $\sigma$ is not the identity then $\sigma \notin Z(S_n)$.

Suppose that $\sigma$ is the identity.

Then $$\big ( \sigma (1), \sigma (2), \dots , \sigma (n)\big ) = \big (1,2,\dots , n\big )$$ So we have that $$\sigma \cdot (1, \ldots, n) = \big (1,\dots , n\big )\cdot \sigma$$

So when $\sigma$ is the identity then $\sigma \in Z(S_n)$.

This means that $Z(S_n)=\{\text{id}\}$.

Is it right? Or could I improve something?

 

[johng1]

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This is just wrong!. Take $\sigma=(1,2,\cdots, n)$. Then the cycle $(2,3,\cdots,n,1)=(1,2,\cdots,n)$. These are cycles, not n tuples.

Here's a complete proof as indicated in my previous post:Suppose $z\in Z(S_n)$ where $n\geq3$. Consider the cyclic decomposition of $z$.
1. Suppose there is a cycle of z of length at least 3, say $c=(i,j,k\cdots)$ and $z=cc_1c_2\cdots c_m$ where each $c_i$ is a cycle not containing either $i$ or $j$. So each $c_i$ commutes with $(i,j)$. Now the centralizer $C$ of $(i,j)$ is a subgroup. So the product $c_1c_2\cdots c_m\in C,\,z\in C$ implies $c=z(c_1c_2\cdots c_m)^{-1}$ commutes with $(i,j)$. But $((i,j)\circ c)(i)=(i,j)(c(i))=(i,j)(j)=i$ whereas $(c\circ (i,j))(i)=c(j)=k$, and $c$ does not commute with $(i,j)$. So every cycle of $z$ has length at most 2.

2. Suppose $z$ contains the product of 2 2 cycles, say $z=(i,j)(k,l)\cdots$. As in 1, $(i,j)(k,l)$ commutes with the 3 cycle $(i,j,k)$, but $((i,j,k)(i,j)(k,l))(i)=k$ and $((i,j)(k,l)(i,j,k))(i)=i$ is a contradiction.

3. Finally then z consists of at most one 2 cycle, say $z=(i,j)$. But as shown in 1, $(i,j)$ does not commute with $(i,j,k)$ for any $k$ with $k\neq i$ and $k\neq j$. So $z$ consists of all 1 cycles, i.e. $z$ is the identity.

 

[evinda]

Suppose $z\in Z(S_n)$ where $n\geq3$. Consider the cyclic decomposition of $z$.
1. Suppose there is a cycle of z of length at least 3, say $c=(i,j,k\cdots)$ and $z=cc_1c_2\cdots c_m$ where each $c_i$ is a cycle not containing either $i$ or $j$. So each $c_i$ commutes with $(i,j)$.

Why does $c_i$ commute with $(i,j)$?

 

[johng1]

Disjoint cycles commute.