# Let $n$ be an odd positive integer and $a\in S_n$ be an $n$-cycle. Show that the order of $C(a)$ must be odd.

I am working on the following problem from group theory:

If $$n$$ is odd and $$a\in S_n$$ is an $$n$$-cycle, $$a=(a_1,a_2,......,a_n)$$, show that no element of the centralizer $$C(a)=\{g\in S_n \mid ga=ag\}$$ of $$a$$ has order $$2$$.

What I did so far is trying to prove $$a(a_i,a_j)a^{-1}$$ is not equal to $$(a_i,a_j)$$, by decomposing the $$a$$ and $$a^{-1}$$, I get $$[(a_1,a_2)(a_2,a_3)..(a_{i-1},a_i)(a_i,a_{i+1})...(a_{j-1},a_j)(a_j,a_{j+1})...(a_{n-1},a_n)](a_i,a_j)[(a_n,a_{n-1})...(a_{j+1},a_j)(a_j,a_{j-1})...(a_{i+1},a_i)(a_i,a_{i-1})...]=(a_{i+1},a_{j+1})$$ which is not equal to $$(a_i,a_j)$$, but I didn't use the condition that $$n$$ is odd.

someone can help me to figure out this problem? Thank you.

• Note that not all elements of order 2 are 2-cycles. In general, they decompose as sets of disjoint 2-cycles, however. It might help to check that when $n=4$ we have that $(1 2 3 4)$ commutes with $(1 3)(2 4)$. – Rolf Hoyer Dec 8 '18 at 22:31
• There's a difference is meaning between "can not" and "cannot". I think the latter is what you intend here. – Shaun Dec 8 '18 at 22:31

$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$$It suffices to compute the conjugacy class of the cycle $$a = (1 2 3 \dots n).$$ It is well known that this is the set of all $$n$$-cycles, of which there are $$\frac{n \cdot (n-1) \cdots 2 \cdot 1}{n}.$$ Therefore, by orbit-stabilizer, the centralizer of $$a$$ has order $$n$$, and thus coincides with $$\Span{a}$$, a group of odd order $$n$$, which thus does not contain elements of order $$2$$.

In this answer, $$n$$ is an arbitrary (not necessarily odd) positive integer. Without loss of generality, we may assume that $$a=(1\;2\;3\;\ldots\;n)$$. Define $$f:S_n\to S_n$$ as the conjugation by $$a$$, namely, $$f(g)=aga^{-1}$$ for all $$g\in S_n$$. Consequently, $$C(a)=\text{Fix}(f)=\big\{g\in S_n\,\big|\,f(g)=g\big\}\,.$$ For $$r\in\mathbb{Z}_{\geq 0}$$, write $$f^r$$ as the $$r$$-time iteration of the function $$f$$, namely, $$f^0:=\text{id}_{S_n}$$, $$f^1:=f$$, $$f^2:=f\circ f$$, $$f^3:=f\circ f\circ f$$, and so on. Finally, $$[l]$$ denotes the set $$\{1,2,\ldots,l\}$$ for every nonnegative integer $$l$$ (here, $$[0]:=\emptyset$$), and $$H:=\langle a\rangle \leq S_n$$ is a cyclic subgroup of $$S_n$$ of order $$n$$.

Let $$b\in C(a)$$. Decompose $$b$$ as a product of disjoint cycles $$s_1s_2\ldots s_k$$ (and without loss of generality, we may suppose that $$1$$ appears in $$s_1$$). Because $$f(b)=b$$, $$f(s_1),f^2(s_1),f^3(s_1),\ldots\in \{s_1,s_2,\ldots,s_k\}\,.$$ Because $$H$$ acts transitively on $$[n]$$, it follows that $$s_1,s_2,\ldots,s_k$$ have the same size, and we may assume without loss of generality that $$s_j=f^{j-1}(s_1)\text{ for }j=1,2,\ldots,k\,.$$ Note also that we must have $$k\mid n$$, so $$n=kq$$ for some positive integer $$q$$.

As an abuse of notation, we write $$t\in s_j$$ if a number $$t\in [n]$$ appears in the cycle $$s_j$$. This proves that $$j\in s_j$$ for all $$j\in [k]$$. Suppose that $$s_1=(t_1\;t_2\;t_3\;\ldots\;t_q)$$ for some $$t_1,t_2,t_3,\ldots,t_q\in [n]$$, with $$t_1:=1$$. Then, $$t_\mu+j-1\in s_j$$ for every $$\mu\in[q]$$ and $$j\in[k]$$, whence $$\{t_1,t_1+1,\ldots,t_1+k-1\},\{t_2,t_2+1,\ldots,t_2+k-1\},\ldots,\{t_q,t_q+1,\ldots,t_q+k-1\}$$ form a partition of $$[n]$$. As a consequence, $$t_\mu\equiv 1\pmod k$$ for every $$\mu\in[q]$$.

Furthermore, by applying $$f$$ on $$s_1$$ for $$k$$ times, we get $$(t_1+k\;t_2+k\;\ldots\;t_q+k)=(t_1\;t_2\;\ldots\;t_q)\,,$$ where the addition is considered modulo $$n$$. If $$t_{\nu+1}=t_1+k$$ for some $$\nu\in[q]$$, then $$t_{\nu+l}=t_{l}+k$$ for $$l\in[q]$$ (where the indices are considered modulo $$q$$). If $$d:=\gcd(\nu,q)> 1$$, then $$t_1+\frac{n}{d}=t_1+\left(\frac{q}{d}\right)k=t_{\left(\frac{q}{d}\right)\nu+1}=t_1\,,$$ which is absurd. Ergo, $$d=1$$. That is, $$t_1,t_2,\ldots,t_q$$ form an arithmetic progression (modulo $$n$$) in $$[n]$$. Therefore, $$t_\mu=1+kr(\mu-1)$$ for some integer $$r\in\{1,2,\ldots,q\}$$ such that $$\gcd(r,q)=1$$.

In other words, fix a positive integer $$k$$ that divide $$n$$ and fix a cycle $$s_1=\big(1\;\;\;1+kr\;\;\;1+2kr\;\;\;\ldots\;\;\;1+(q-1)kr\big)\,,$$ where $$q=\dfrac{n}{k}$$ as before and $$r\in[q]$$ is coprime to $$q$$. (There will be $$\varphi(q)=\varphi\left(\dfrac{n}{k}\right)$$ possible choices of $$s_1$$, where $$\varphi$$ is Euler's totient function.) Then, $$b=s_1s_2\cdots s_k=s_1\,f(s_1)\,f^2(s_1)\,\cdots\,f^{k-1}(s_1)$$ is equal to $$a^{kr}$$. This shows that $$b\in H$$. Thus, $$C(a)=H=\langle a\rangle$$.

In particular, if $$n$$ is odd, then $$C(a)$$ is of an odd order, and no element of $$C(a)$$ has an even order. (You do not need to know completely what $$C(a)$$ contains to show that no element of $$C(a)$$ has an even order, given that $$n$$ is odd. Somewhere earlier in this answer, which I leave it as a mystery, already gives you a proof of that statement.) As a side note, this provides a different (albeit long and ineffective) proof that $$n=|H|=\sum_{\substack{k\in[n]\\{k\mid n}}}\,\varphi\left(\frac{n}{k}\right)=\sum_{\substack{q\in[n]\\{q\mid n}}}\,\varphi\left(q\right)\,.$$

With great insights from Andreas Caranti's answer, I have found the following result. Let $$a\in S_n$$ be arbitrary. Suppose that the decomposition of $$a$$ into a product of disjoint cycles is $$\prod_{\ell \in \mathbb{Z}_{>0}}\,\prod_{i=1}^{m_\ell}\,\sigma_{\ell,i}\,,$$ where $$m_\ell\in\mathbb{Z}_{\geq 0}$$ is the number of $$\ell$$-cycles in this cycle decomposition of $$a$$ and, for $$i\in [m_\ell]$$, $$\sigma_{\ell,i}$$ is a cycle in $$S_n$$ of length $$\ell\in\mathbb{Z}_{>0}$$.

Then, the centralizer of $$a$$ is the internal direct product $$C(a)=\prod_{\ell\in\mathbb{Z}_{>0}}\,G_\ell\,,$$ where $$G_\ell$$ is a subgroup of $$S_n$$ that stabilizes $$\tau_\ell:=\displaystyle \prod_{i=1}^{m_\ell}\,\sigma_{\ell,i}$$ for all $$\ell\in\mathbb{Z}_{>0}$$ and fixes every element of $$[n]$$ not appearing in $$\sigma_{\ell,i}$$ for all $$i\in[m_\ell]$$. This subgroup $$G_\ell$$ is contains the subgroup $$H_\ell$$ generated by $$\sigma_{i,\ell}$$ for $$i\in[m_\ell]$$ as a normal subgroup. Note that $$H_\ell\cong (Z_\ell)^{m_\ell}$$, where $$Z_\ell$$ is the cyclic group of order $$\ell$$.

Write each $$\sigma_{i,\ell}$$ as $$\left(t_{i,\ell}^1\;\;\;t_{i,\ell}^2\;\;\;\ldots\;\;\;t_{i,\ell}^\ell\right)\,,$$ where $$t_{i,\ell}^\mu\in [n]$$ for all $$\mu\in[\ell]$$ and $$t_{i,\ell}^1$$ is the smallest among these $$t_{i,\ell}^\mu$$. Let $$K_\ell$$ be the subgroup of $$G_\ell$$ isomorphic to the symmetric group $$S_{m_\ell}$$ such that the elements of $$K_\ell$$ are of the form $$\zeta\in S_n$$ such that $$\zeta\left(t_{i,\ell}^\mu\right)=t_{\delta(i),\ell}^\mu$$ for some $$\delta\in S_{m_\ell}$$, and for all $$i\in[m_\ell]$$ and $$\mu\in[\ell]$$, and $$\zeta$$ fixes all other elements of $$[n]$$. Then, $$G_\ell$$ is the internal semidirect product $$H_\ell\rtimes K_\ell$$.

Consequently, $$C(a)$$ is the subgroup $$\prod_{\ell\in\mathbb{Z}_{>0}}\,\left(H_\ell \rtimes K_\ell\right)\cong \prod_{\ell\in\mathbb{Z}_{>0}}\,\Big(\left(Z_\ell\right)^{m_\ell}\rtimes S_{m_\ell}\Big)\,.$$ This subgroup is of order $$\prod_{\ell\in\mathbb{Z}_{>0}}\,\left(\ell^{m_\ell}\,m_\ell!\right)\,.$$

Let $$n=2m+1$$. Suppose on the contrary there is $$g$$ of order 2 in $$C(a)$$. Then $$a=gag^{-1}=gag$$. It follows that $$a^2=(gag)(gag)=ga^2g=ga^2g^{-1}$$.

Note that since $$n=2m+1$$ then $$a^2=(a_1\quad a_3\quad \ldots \quad a_{2m+1}\quad a_2\quad a_4\quad \ldots \quad a_{2m})$$ and $$a^2=ga^2g^{-1}=(g(a_1)\quad g(a_3)\quad \cdots g(a_{2m+1})\quad g(a_2)\quad g(a_4)\quad \cdots \quad g(a_{2m}))$$.

From the two expressions of $$a^2$$ above the value of $$g$$ is determined by $$g(a_1)$$. Note that taken modulo $$n=2m+1$$ the even indices $$2,4,\ldots,2m$$ can be written as $$2m+3,2m+5,\ldots, 4m+1$$.

Then we can assume that $$g(a_1)=a_{2k+1}$$ where $$0\leq k\leq 2m$$. It follows from the two representation of $$a^2$$ above that $$g(a_{2k+1})=a_{4k+1}$$. But since $$g$$ is of order 2 we also have $$g(a_{2k+1})=a_1$$. Hence $$a_1=a_{4k+1}$$ which can only happen if $$1\equiv 4k+1\pmod n$$, that is when $$n\mid 4k$$. Since $$n$$ is odd, $$n$$ and 4 are relative prime. Hence $$n\mid k$$ that is $$2m+1\mid k$$.

But since $$0\leq k\leq 2m$$ then $$k=0$$. Which means that $$g(a_1)=a_1$$ from which it follows that $$g(a_i)=a_i$$. So $$g$$ is the identity map (contradiction).