Let $a$ be an element of a group $G$ and $|a| = 7.$ Show $a$ is the cube of some element of $G$. Having some trouble understanding how to proceed w/ this hw question:

Let $a$ be an element of a group $G$ and $|a| = 7.$ Show $a$ is the cube of some element of $G$.

Attempt:
$a^7 = e$ by hypothesis.
So need to show $a = b^3$ for some $b \in G$
$a^7 = (b^3)^7 = b^{21} = e$
But is this even true? Have I actually shown something? 
 A: We have $$\begin{align}a^{15}&=a^{14}\cdot a \\ &=(a^7)^2\cdot a \\ &=e\cdot a \\ &=a,\end{align}$$ so that $a=b^3$ with $b=a^5\in G$.
A: There could be a $b: |b| = 21; b^3 =a$ but you know nothing about the group and other elements.  In fact the group could be simply $G=<a>$ cyclic group. 
So if this statement is always true it must be always true for $G = <a>$ and we have some $(a^k)^3 = a$.
Is that possible? is it always possible?
But if we have $(a^k)^3 = a^{3k} = a$ then we have $a = a^1= a^1*e^w = a^1*(a^7)^w = a^{7w+1}$.
If we can find a $k$ so that $3k = 7w+1$ for some $w$ we'd be done.... Because $(a^k)^3 = a^{3k} = a^{7w+1}=a*(a^7)^w = a$.
And as $3$ and $7$ are relatively prime we know we can find such a $k$.
A: Is $a$ the cube of some power of $a$?  There are only six possible powers to check, so directly checking them is pretty quick...
A: Let $b=a^{5}$, then $b^{3}=(a^{5})^{3}=a^{15}=(a^{7})^{2}a=ea=a$.
A: Let's consider the subgroup $H$ of order $7$ generated by $a$ and consider the function $\varphi:H \to H$ defined by $\varphi(x)=x^3$.  Because $H$ is abelian, it's easy to show that $\varphi$ is in fact a homomorphism.  Show that $\ker \varphi = \{e \},$ which means that $\varphi$ is in fact an isomorphism, so (since $H$ is finite) it must be surjective.  And that, in turn, means that every element of $H$ must be the cube of some other element of $H$.
A: This answer augments the other answers, since $3\times 5=15=2(7)+1$, by proving a slightly more general result.

Theorem: Let $a\in G$ for some group $G$. If there exist $h, k\in\Bbb N$ such that $hk\equiv 1\pmod{\lvert a\rvert}$ (i.e., there is some $\ell\in\Bbb Z$ with $hk=1+\ell\lvert a\rvert$), then $a$ has an $h$th root, $a^k$.

Proof: We have $1+\ell\lvert a\rvert=kh$ for some $\ell\in\Bbb Z$, so that $$\begin{align}a^{hk}&=a^{\ell\lvert a\rvert}\cdot a \\ &=(a^{\lvert a\rvert})^\ell\cdot a \\ &=e\cdot a \\ &=a,\end{align}$$ so that then $(a^k)^h=a$.$\square$
A: Another (equivalent) way:
$$|a|=7\implies a^7=e\iff a\cdot a^6=e$$
Then multiplying both sides on the right by $(a^{-1})^6=a^{-6}$: $$a\cdot a^6\cdot a^{-6}=e\cdot a^{-6}=a^{-6}$$
So $$ae=a=a^{-6}=\left((a^{-1})^2\right)^3$$
where $b=(a^{-1})^2$ is an element of the group by existence of inverses and closure. Thus $a=b^3$.
