Take $G$ to be a group of order $600$. Prove that for any element $a$ $\in$ G there exist an element $b$ $\in$ G such that $a = b^7$.
My thought process: since $a = b^7$ $\implies$ $|a| = |b^7|$. Consequently $\operatorname{lcm}(1,|a|) = \dfrac{1}{7}\operatorname{lcm}(7,|b|)$ implies $7|a| = 7|b|$ so $|a| = |b|$. I don't know where I would go from here or if this is even the right approach.
Take the cyclic subgroup generated by $a$. It has some order $k$ so $a^{1+kn}=a$ for all $n$. Let $b=a^l$
$$ b^7=a^{1+kn}\\ a^{7l}=a^{1+kn}\\ $$
$7l \equiv 1 \; (mod \; k)$
Suppose $k=2$, then $l=1$ works and $b=a$ satisfies $b^7=b=a$.
Suppose $k=20$, then $l=3$ work $b=a^3$ satisfies $b^7=a^{21}=a$
What are the possible $k$ and what can you then say about the solvability of $7l \equiv 1$?
Hint: $\gcd(600,7)=1 \implies 1=600m+7n$. Indeed, $ 1 = 3 \cdot 600 - 257 \cdot 7$.
Proposition Let $G$ be a finite group and $n$ a positive integer. Then the map $f: G \mapsto G$ defined by $f(g)=g^n$ is a bijection if and only if gcd$(|G|,n)=1$.
Proof (sketch) Bézout yields $1=k|G|+mn$, for some integers $k, m$. Then $g=g^{k|G|+mn}=g^{mn}$. Hence if $g^n=h^n$, then $g=g^{mn}=h^{mn}=h$. So $f$ is injective and since $G$ is finite it must be bijective. Conversely, assume gcd$(n,|G|)\neq 1$. Then we can find a prime $p$ with $p \mid n$ and $p \mid |G|$. By Cauchy's Theorem there is a non-trivial $g \in G$ with order$(g)=p$. Then $g^n=g^{p \cdot \frac{n}{p}}=1^\frac{n}{p}=1=1^n$. Since $f$ is injective this yields $g=1$, a contradiction.
First I show how to reduce it to computing $\,7^{-1}\!\pmod{\!600};$ then I explain why this reduction works so universally due to persistence of GCDs having linear (Bezout) representation.
$\langle a\rangle$ has order $\,n\mid 600\,$ so $\langle a\rangle$ is an image of $\,\Bbb Z/600,\,$ so it suffices to solve it mod $600;\,$ said more concretely $\,a\equiv 7b\pmod{\!600\!=\!nk}\,\Rightarrow\,a\equiv 7b\pmod{\!n}.$ $\,7^{-1}$ exists by $(7,600)=1;\,$ explicitly
$\!\bmod 600\!:\ 7b\equiv a\iff b\equiv a/7\equiv\color{#c00}{343}\,a,\ $ by $ \underbrace{\ 7^{\large 4}\equiv 1\,\Rightarrow\, 1/7\equiv 7^{\large 3}\!\equiv \color{#c00}{343}}_{\Large\ \ 7^{\LARGE 2}\, \equiv\ \pm1\ \bmod\, 24\ \,\&\, \ 25\qquad}$
Remark $\ $ The proof requires only the coprimality of $\,k,n = 7,600\,$ since then we have a Bezout equation $\, k'k - n' n = 1\,$ $\,\Rightarrow\,\bmod n\!:\ k'\equiv k^{-1} =: (1/k)_n\,$ so we can proceed as above, i.e.
$$\begin{align} \color{#c00}{na = 0}\,\Rightarrow&\,\ k'(ka) = (1\!+\!n'n)a = a\! +\! n'(\color{#c00}{na}) = a\ \quad\rm Additive\ Form\\[.4em] \color{#c00}{a^{\large n} = 1}\,\Rightarrow&\,\ (a^{\large k})^{\large k'} = a^{\large 1\ +\ n'n}\ \ \ =\ \ a\,(\color{#c00}{a^{\large n}})^{\large n'} =\ a\ \quad\rm Multiplicative\ Form\end{align}\quad$$
hence $\quad\ \ \bbox[5px,border:1px solid #c00]{\begin{align} \text{if $\,\ n\,a = 0 = n\,b\ $ then }\ kb = a &\iff b = (1/k)_n\, a\\[.4em] \text{if $\:\!\ a^n\, =\, 1\, =\, b^n \ $ then }\ b^{\large k} = a &\iff b = a^{\large (1/k)_n} \end{align}}$
This works so nicely because whenever a gcd has a linear representation (Bezout) it is universal, i.e. the gcd persists in extension rings /algebras / modules - just as above
Note $\ $ Alternatively we can invert $7$ mod $\ 600 = 24\cdot 25\ $ using Easy CRT as below
$\!\bmod 24\!:\ \dfrac{1}7\ \equiv\ \dfrac{49}7\ \equiv\ 7\qquad\qquad\qquad\ $
$\!\bmod 25\!:\ \dfrac{1}7\equiv \dfrac{-49}7\equiv-7\ \ {\rm so}\ \smash{\ \dfrac{1}7\equiv \overbrace{-7\!+\!25\left[\dfrac{14}{25}\bmod 24\right]}^{\text{by Easy CRT}}\! \equiv -7\!+\!25[14]\equiv \color{#c00}{343}\pmod{600}}$
