$k$ cosets of normal subgroup and $k$th power of group element

Question

I wrote a proof, but I'm afraid it's not quite perfect.

If one wants to prove that for a finite group $G$, where $N \trianglelefteq G$, and $|G:N|=k$, then $a^k \in N$ for all $a \in G$, he can proceed as follows:

Since $N \trianglelefteq G$, $G/N$ is a group, and there exists some homomorphism $\phi$ from $G/N$ to some group, such that $\ker \phi=N$. The fibers under this homomorphism are classes $\bar{a}_i$ ($1\leq i \leq k$). For any $a \in G$, since there are only $k$ cosets of $N$ in $G$, cosets $a^rH$ ($0 \leq r \leq (k-1)$), such that powers of $a$ represent elements from distinct equivalence classes $\bar{a}$, $a^k$ must be in $N$.

I know that my proof is flawed, but I think I might have the right idea, and I would appreciate some hints.

Answer 1

Consider the next Lemma:

Let $G$ be a finite group with $|G|=m$. Let $b\in G$ be any element $\neq e$ of $G$. Then we have $b^m=e$.

Proof:

Take $\langle a\rangle$ the subgroup generated $b$. Set-wise we get the subset $\{b,b^2,...,b^{r-1},b^r=e\}$ the finite powers of $b$ in $G$. By Lagrange's theorem $r\mid m$, so $m=r\mu$ for some integer $\mu$.

So, $b^m=\left(b^r\right)^{\mu}=e^{\mu}=e$,

$\Box$

Now for your problem you know that for $aN\in G/N$ we have $(aN)^k=eN$. Then $a^kN=eN$, so $a^k\in N$.