If $a^3=e$, then $a$ has a square root Assuming $a\in G$ where $G$ is a group.
I'm not sure why this is hard for me. Essentially, the problem is just saying:

If $a^3=e$, then there exists $x \in G$ such that $a=x^2$.

Can somebody give me a hint or a direction to start in? (not full solution, please--I want to figure it out myself)
Thank you.
 A: Hint: Multiply by $a$ to find
$$a^4 = a$$
A: Hint $\,\ a^{\large 3}= 1\,\Rightarrow\,a^{\large 3n}=1\,\overset{\times\, a}\Rightarrow\,a^{\large 1+3n} = a\, $ is a square if $\ 2\mid 1\!+\!3n,\,$ e.g. $\, n = \,\ldots\ $ QED
Remark $ $ Here is the intuition.  $\ a^{\large 3}= 1\ $ implies that exponents on $\,a\,$ can be considered mod $\,3,\,$ 
$$\color{#c00}{a^{\large 3}= 1}\ \Rightarrow\ a^{\large k+3n} = a^{\large k} (\color{#c00}{a^{\large 3}})^{\large n} = a^{\large k}\ \ \ {\rm so}\ \ \ a^{\large j}\! = a^{\large j\ {\rm mod}\ 3} $$
Therefore we can replace the exponent $1$ in $\,a^1\,$ by any $\,j\equiv 1\pmod 3,\,$ which includes $ $ even $\,j,\,$ i.e. $\,{\rm mod}\ 3,\,$ we have that $1$ is "even", i.e. $\ 2\mid 1,\,$  i.e. $2$ is invertible. This generalizes as follows.

If $\,\color{#c00}{a^k = 1}\,$ and $\,\gcd(n,k)=1\,$ then $\,a\,$ is an $n$'th power. Indeed, by above it suffices to find a multiple $\,jn\,$ of $\,n\,$ that is $\,\equiv 1\pmod k,\,$ i.e. to invert $\,n\,$ mod $\,k,\,$ which is easy:
$\qquad\qquad$ by Bezout, there are $\,i,j\in\Bbb Z\,$ with $\ jn = 1 + ik\ $ so $\ (a^j)^n = a(\color{#c00}{a^k})^i = a$
Note how the problem reduces to the problem of division mod $\,k.\,$ The structure underlying this reduction will become clearer when one studies cyclic groups and modules (over $\,\Bbb Z/k)$.
A: Hint: Try $x$ as a power of $a$ (since these are the only things being assuredly in $G$).
A: Heres another way to view it simply:
$$ a^3=e $$
$$ aaa=e      $$ 
$$   (aa)a=e    $$
$$       (a^2)a=e$$
$$      (a^2)a(a^{-1})=e(a^{-1})$$
$$   a^2(aa^{-1})=e(a^{-1})   $$
$$       a^2=a^{-1}$$
$$       $$
$$       $$
Where if you have defined $a \in G $ will imply $a^{-1} \in G$ by the definition of a group
