Csdrhrt

I am working on the following problem from group theory:

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

$newcommand{Span}[1]{leftlangle #1 rightrangle}$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 $f:S_nto S_n$ as the conjugation by $a$, namely, $f(g)=aga^{-1}$ for all $gin S_n$. Consequently, $$C(a)=text{Fix}(f)=big{gin S_n,big|,f(g)=gbig},.$$
For $rinmathbb{Z}_{geq 0}$, write $f^r$ as the $r$-time iteration of the function $f$, namely, $f^0:=text{id}_{S_n}$, $f^1:=f$, $f^2:=fcirc f$, $f^3:=fcirc fcirc f$, and so on. Finally, $[l]$ denotes the set ${1,2,ldots,l}$ for every nonnegative integer $l$ (here, $[0]:=emptyset$), and $H:=langle arangle leq S_n$ is a cyclic subgroup of $S_n$ of order $n$.

Let $bin C(a)$. Decompose $b$ as a product of disjoint cycles $s_1s_2ldots s_k$ (and without loss of generality, we may suppose that $1$ appears in $s_1$). Because $f(b)=b$, $$f(s_1),f^2(s_1),f^3(s_1),ldotsin {s_1,s_2,ldots,s_k},.$$
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 $kmid n$, so $n=kq$ for some positive integer $q$.

As an abuse of notation, we write $tin s_j$ if a number $tin [n]$ appears in the cycle $s_j$. This proves that $jin s_j$ for all $jin [k]$. Suppose that $s_1=(t_1;t_2;t_3;ldots;t_q)$ for some $t_1,t_2,t_3,ldots,t_qin [n]$, with $t_1:=1$. Then, $t_mu+j-1in s_j$ for every $muin[q]$ and $jin[k]$, 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, $t_muequiv 1pmod k$ for every $muin[q]$.

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 $nuin[q]$, then
$$t_{nu+l}=t_{l}+k$$
for $lin[q]$ (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 $rin{1,2,ldots,q}$ such that $gcd(r,q)=1$.

In other words, fix a positive integer $k$ that divide $n$ and fix a cycle $$s_1=big(1;;;1+kr;;;1+2kr;;;ldots;;;1+(q-1)krbig),,$$ where $q=dfrac{n}{k}$ as before and $rin[q]$ is coprime to $q$. (There will be $varphi(q)=varphileft(dfrac{n}{k}right)$ possible choices of $s_1$, where $varphi$ is Euler's totient function.) Then, $$b=s_1s_2cdots s_k=s_1,f(s_1),f^2(s_1),cdots,f^{k-1}(s_1)$$ is equal to $a^{kr}$. This shows that $bin H$. Thus, $C(a)=H=langle arangle$.

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 $$n=|H|=sum_{substack{kin[n]\{kmid n}}},varphileft(frac{n}{k}right)=sum_{substack{qin[n]\{qmid n}}},varphileft(qright),.$$

With great insights from Andreas Caranti's answer, I have found the following result. Let $ain S_n$ 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 $m_ellinmathbb{Z}_{geq 0}$ is the number of $ell$-cycles in this cycle decomposition of $a$ and, for $iin [m_ell]$, $sigma_{ell,i}$ is a cycle in $S_n$ of length $ellinmathbb{Z}_{>0}$.

Then, the centralizer of $a$ is the internal direct product
$$C(a)=prod_{ellinmathbb{Z}_{>0}},G_ell,,$$
where $G_ell$ is a subgroup of $S_n$ that stabilizes $tau_ell:=displaystyle prod_{i=1}^{m_ell},sigma_{ell,i}$ for all $ellinmathbb{Z}_{>0}$ and fixes every element of $[n]$ not appearing in $sigma_{ell,i}$ for all $iin[m_ell]$. This subgroup $G_ell$ is contains the subgroup $H_ell$ generated by $sigma_{i,ell}$ for $iin[m_ell]$ as a normal subgroup. Note that $H_ellcong (Z_ell)^{m_ell}$, where $Z_ell$ is the cyclic group of order $ell$.

Write each $sigma_{i,ell}$ as
$$left(t_{i,ell}^1;;;t_{i,ell}^2;;;ldots;;;t_{i,ell}^ellright),,$$
where $t_{i,ell}^muin [n]$ for all $muin[ell]$ 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 $zetain S_n$ such that
$$zetaleft(t_{i,ell}^muright)=t_{delta(i),ell}^mu$$
for some $deltain S_{m_ell}$, and for all $iin[m_ell]$ and $muin[ell]$, and $zeta$ fixes all other elements of $[n]$. Then, $G_ell$ is the internal semidirect product $H_ellrtimes K_ell$.

Consequently, $C(a)$ is the subgroup
$$prod_{ellinmathbb{Z}_{>0}},left(H_ell rtimes K_ellright)cong prod_{ellinmathbb{Z}_{>0}},Big(left(Z_ellright)^{m_ell}rtimes S_{m_ell}Big),.$$
This subgroup is of order $$prod_{ellinmathbb{Z}_{>0}},left(ell^{m_ell},m_ell!right),.$$

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 $$a^2=(a_1quad a_3quad ldots quad a_{2m+1}quad a_2quad a_4quad ldots quad a_{2m})$$ 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 $0leq kleq 2m$. 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 $1equiv 4k+1pmod n$, that is when $nmid 4k$. Since $n$ is odd, $n$ and 4 are relative prime. Hence $nmid k$ that is $2m+1mid k$.

But since $0leq kleq 2m$ 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).