I have in some notes, this statement:
Given $C_3\ltimes C_7$ we know that for $a\in C_3$ and $b\in C_7$, and some $k$:
$$aba^{-1}=b^k$$
$$k^3\equiv 1(7)$$
The reason given is that $a^3=1$. Unfortunately, I'm stuck on making the connection between that fact and the conclusion that $k^3\equiv 1(7)$. I'd be grateful for some clarification.
Observe: $$\begin{eqnarray*} aba^{-1}&=&b^k\\ a\left(aba^{-1}\right)a^{-1}&=&ab^ka^{-1}\\ a^2ba^{-2}&=&ab^ka^{-1}\\ &=& a b b b \cdots b a^{-1} \\ &=&ab\; \left(a^{-1}a\right)\; b \;\left(a^{-1}a\right)\;b \cdots b \; \left(a^{-1}a\right)\;b a^{-1} \\ &=&\left(ab a^{-1} \right)\;\;\left(ab a^{-1} \right)\;\left(ab a^{-1} \right)\;\cdots \left(ab a^{-1} \right)\; \\ &=& \left(b^k\right)\left(b^k\right)\left(b^k\right)\cdots \left(b^k\right)\\ &=& \left(b^k\right)^k\\ &=& b^{k^2}\\ \end{eqnarray*}$$ You can see what will happen if we do it again: $a^3ba^{-3}=b^{k^3}$. But since $a^3=a^{-3}=\operatorname{id}$, this gives us that $b^{k^3}=b$. Since the order of $b$ is $7$, that means we can say that the exponents of left and right hand sides are congruent modulo $7$. Thus, $$k^3 \equiv 1 \pmod 7$$