Given $o(a)=5$, prove $C(a)=C(a^{3})$

Question

Given $o(a)=5$, prove $C(a)=C(a^{3})$

At this point I would like a hint rather than a full solution. I know we are given $a^{5}=e$ and that we wish to prove this implies that $C(a) =\{ x \in G:xa=ax \}=\{x \in G : xa^{3}=a^{3}x \} =C(a^{3})$. However, I can't around as to how exactly I am going to show the latter.

I take a stab at it:

\begin{align} ax&=eax \\ &=a^{5}ax \tag{$o(a)=5$} \\ &=(aaaaa)ax \tag{Definition of power} \\ &=aaa(aaax) \tag{Associativity} \\ &=aaa(xaaa) \tag{Assumption}\\ &=(aaax)aaa \tag{Associativity}\\ &=(xaaa)aaa \tag{Assumption} \\ &=xaa^{5} \tag{Associativity}\\ &=xae \tag{$o(a)=5$} \\ &=xa \tag{Identity} \end{align}

Answer 1

The standard way of showing set equality is to take an element of one set and show that is also belongs to the other set. Then go the other way.

I'll take the easy one, showing that $C(a) \subseteq C(a^3)$.

Let $x \in C(a)$. That means that $xa = ax$. But then, since the group operation is associative, we have $$ a^3x = aaax = aa(ax) = aa(xa) = a(ax)a = a(xa)a = (ax)aa = (xa)aa = xa^3 $$ so $x\in C(a^3)$.

Going the other way (i.e. showing that $C(a^3) \subseteq C(a)$) requires a small trick, but it goes along the same lines.

Answer 2

HINT:

$a^5 = e$ so $a = a^{5k+1}$, therefore

$$a = (a^3)^2$$

Answer 3

Try this roadmap:

• $C(a)=C(\langle a \rangle)$

• $ \langle a \rangle = \langle a^3 \rangle$

More generally, $C(a)=C(a^k)$ for all $k$ with $\gcd(k,o(a))=1$ because $ \langle a \rangle = \langle a^k \rangle$ in this case.