Find an element $a$ from some group such that $|a|=6$ and $C(a) \neq C(a^3)$ Find an element $a$ from some group $G$ such that $|a|=6$ and $C(a) \neq C(a^3)$.
$|a|$ is the order of $a$.
$C(a)$ is the Centraliser of $a$ in $G$.
This question is from Contemprary Abstract Algebra by Joseph A Gallian 4th edition,
Chapter 3, Question 24.
 A: We want $a$ to have order $6$? Let's try $a=(1\ 2\ 3\ 4\ 5\ 6)$ in the group $S_6$. Let's see, $a^3=(1\ 4)(2\ 5)(3\ 6)$ right? Does $C(a)$ mean the set of group elements that commute with $a$? So we're looking for something that commutes with $a^3$ but not with $a$? Hmm. The transposition $(1\ 4)$ commutes with $a^3$, doesn't it? I wonder if it commutes with $a$.
A: The simplest example that i see is the following. Take
$$G=S_5,$$ 
where $S_5$ denotes the group of all the permutations of the set $\{1,2,3,4,5\}$, and take 
$$a=(123)(45).$$
So we have that 
$$a^3=(45).$$
The centralizer of $a$ is given by
$$ C(a)=\left\{\sigma\tau \;| \; \sigma \in \langle(123)\rangle, \tau \in \langle(45)\rangle \right\} \simeq \mathbb Z_3 \times\mathbb Z_2,$$
while the centralizer of $a^3$
$$C(a^3)=\left\{\sigma\tau \;| \; \sigma \in S_3, \tau \in \langle(45)\rangle \right\} \simeq S_3 \times \mathbb Z_2.$$
A: Since $|a| =6$ must divide the group's order, and since the two groups of order $6$ do not work here, the smallest example must have at least $6n$ elements, with $n \geq 2$. But it turns out that the Dihedral group $D_6 = \langle a,s \,|\, a^6=s^2=1, \, as=sa^{-1} \rangle$ of order $12$ works. It is routinely checked that
(1) $a,a^2,a^3,a^4,a^5,a^6=1$ are distinct elements;
(2) $as \neq sa \implies s \notin C(a)$;
(3) $a^3s=sa^{-3} = sa^3 \implies s \in C(a)$.
Therefore $|a|=6$ by (1) and $C(a) \neq C(a^3)$ by (2) & (3).
