Time taken by KaTeX to render formulæ : 69 ms PDF version

Theorem

If gg is an element of G=SnG=S_n then CG(g)=gC_G(g)=\langle g\rangle if and only if the cycles of gg are of unequal coprime lengths.
Alternatively, CG(g)=gC_G(g)=\langle g\rangle if and only if gg has cycles of coprime length with at most one 1-cycle.


Proof. Since powers of gg centralise gg it follows that gCG(g)\langle g\rangle\subseteq C_G(g) so
CG(g)=g    g=CG(g)    g=CG(g)\begin{aligned} C_G(g)=\langle g\rangle\iff \lvert\langle g\rangle\rvert=\lvert C_G(g)\rvert\iff\lvert g\rvert=\lvert C_G(g)\rvert \end{aligned}
(1) Let GG act on itself by conjugation. Then xgx1=x    xg=gxxgx^{-1} = x\iff xg=gx so
Stab(g)=CG(g)\begin{aligned} \text{Stab}(g)=C_G(g) \end{aligned}
(2)
Orb(g)=the number of distinct conjugates of g=the number of permutations with the same cycle structure as g\begin{aligned} \lvert \text{Orb}(g)\rvert&=\text{\small the number of distinct conjugates of g}\\ &=\text{\small the number of permutations with the same cycle structure as }g \end{aligned}
Let aa be the product of the lengths of the cycles of gg
Let bb be the number of cycles of equal length.
Let ll be the least common multiple of the lengths of the cycles, g=la\lvert g\rvert=l\leq a.

Then the number of permutations with the same cycle structure as gg is
n!ab=Gab\begin{aligned} \frac{n!}{ab}=\frac{\lvert G\rvert}{ab} \end{aligned}
Hence
Orb(g)=Gab\begin{aligned} \lvert \text{Orb}(g)\rvert=\frac{\lvert G\rvert}{ab} \end{aligned}
(3) By the Orbit-Stabiliser theorem, Orb(g)×Stab(g)=G\lvert \text{Orb}(g)\rvert\times\lvert \text{Stab}(g)\rvert=\lvert G\rvert so by (3)
Stab(g)=GOrb(g)=GG/ab=abgbg=l\begin{aligned} \lvert \text{Stab}(g)\rvert=\frac{\lvert G\rvert}{\lvert \text{Orb}(g)\rvert}=\frac{\lvert G\rvert}{\lvert G\rvert/ab}=ab\geq\lvert g\rvert b\geq\lvert g\rvert=l \end{aligned}
and it follows that
Stab(g)=g    a=l and b=1\begin{aligned} \lvert \text{Stab}(g)\rvert=\lvert g \rvert\iff a = l \text{ and } b = 1 \end{aligned}
(4) But
a=l    g has coprime cycle lengthsb=1    g has unequal cycle lengths\begin{aligned} a=l\iff g\text{ has coprime cycle lengths}\\ b=1 \iff g \text{ has unequal cycle lengths} \end{aligned}
(5) (6) The result now follows from (1), (2), (4), (5) and (6).
CG(g)=g    Stab(g)=g    the cycles of g are of unequal coprime lengths C_G(g)=\langle g\rangle\iff\lvert \text{Stab}(g)\rvert=\lvert g \rvert\iff\text{the cycles of g are of unequal coprime lengths}
Cycles of coprime length will have unequal lengths unless they are 1-cycles. Hence we may replace unequal by at most one 1-cycle. \blacksquare