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Theorem

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


Proof. Since powers of g centralise g it follows that \langle g\rangle\subseteq C_G(g) so
\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 G act on itself by conjugation. Then xgx^{-1} = x\iff xg=gx so
\begin{aligned} \text{Stab}(g)=C_G(g) \end{aligned}
(2)
\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 a be the product of the lengths of the cycles of g
Let b be the number of cycles of equal length.
Let l be the least common multiple of the lengths of the cycles, \lvert g\rvert=l\leq a.

Then the number of permutations with the same cycle structure as g is
\begin{aligned} \frac{n!}{ab}=\frac{\lvert G\rvert}{ab} \end{aligned}
Hence
\begin{aligned} \lvert \text{Orb}(g)\rvert=\frac{\lvert G\rvert}{ab} \end{aligned}
(3) By the Orbit-Stabiliser theorem, \lvert \text{Orb}(g)\rvert\times\lvert \text{Stab}(g)\rvert=\lvert G\rvert so by (3)
\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
\begin{aligned} \lvert \text{Stab}(g)\rvert=\lvert g \rvert\iff a = l \text{ and } b = 1 \end{aligned}
(4) But
\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).
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