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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.

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  • $\begingroup$ 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 $(1 3)(2 4)$. $\endgroup$ – Rolf Hoyer Dec 8 '18 at 22:31
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    $\begingroup$ There's a difference is meaning between "can not" and "cannot". I think the latter is what you intend here. $\endgroup$ – Shaun Dec 8 '18 at 22:31
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$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$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 $2$.

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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 $$\prod_{\ell\in\mathbb{Z}_{>0}}\,\left(\ell^{m_\ell}\,m_\ell!\right)\,.$$

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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).

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