If $a^3=e$, then $a$ has a square root.

Question

If $a^3=e$, then $a$ has a square root.

I have that $a^3=aaa=e$ and applying $a^{-1}$ twice on both sides of the equation gives us $a=a^{-1}a^{-1}$ and so $a=(a^{-1})^2$. Thus $a$ has a square root.

Is this correct? Is there another way to do this?

Answer 1

Your attempt is correct, and as mentioned there is the other proof, that $a^3 = e$ implies $a^4 =(a^2)^2 =a$.

A more general statement is true, and is easy to prove.

If $G$ is a finite group of odd order then every element of $G$ has a unique square root.

Proof : Let $|G| = 2n-1$, then if $a = g^2 = h^2$ for two elements $g,h$, then $(g^{2})^{n} = (h^2)^n$ implies $g^{2n} = h^{2n}$ implies $g=h$ by Lagrange's theorem. Thus, $a$ has a unique square root.

Finally, the map $S : G \to G$ given by $S(x) = x^2$ is an injective map between finite sets of the same size, hence a surjective map. The result follows.

---

To apply this to your question, if $a^3 = e$ then the cyclic group generated by $a$ has three elements , and you apply the above statement to conclude.

(TASK : Prove that if $G$ is a group such that every element of $G$ has odd order, then every element of $G$ has a square root, using the method above).