$G$ is a group, $N$ is a subgroup of $G$ and $[G:N]=p$ is a prime.
We know if $N$ is normal we can write $G=\bigcup_{i=0}^{p-1}g^{i}N$.
Because $G/N$ is cyclic and $G/N=\langle \overline{g}\rangle$. So any $h\in G$ we can have $\overline{h}=\overline{g}^{k}$. So $h\in g^{k}N,0\leq k\leq p-1$. So $G=\bigcup_{i=0}^{p-1}g^{i}N$.
But is this right when $N$ is not normal？Can we still find a $g$ to make $G=\bigcup_{i=0}^{p-1}g^{i}N$?
I think this may not be right because when $N$ is not normal, all the left coset of $N$ will not form a group so $d=\min\{k\in \mathbb{N^{*}}|g^{k}\in N\}$ may not divide $p$.
Any help will be thanked.
Yes, you can always find a $g$ with that property, though you are correct that the argument you used for normal subgroups will not work in general.
First, consider the case that $G$ is finite. Let $H$ be a subgroup such that $[G:H]=p$ is prime. (I’m using $H$ because I don’t like to use $N$ to denote subgroups that are not Normal. Note that $|G|=p|H|$.
In particular, no Sylow $p$-subgroup of $G$ can be contained in $H$. Let $P$ be a Sylow $p$-subgroup of $G$; then $P\cap H$ is a proper subgroup of $P$, so it is contained in a maximal subgroup $M$ of $P$. However, maximal subgroups of finite $p$-groups are normal, so we know there exists $g\in P$ such that $P=\cup_{i=0}^{p-1} g^iM$. In particular, $g^i\in M$ if and only if $p|i$, and hence $g^i\in P\cap H$ only if $p|i$, hence $g^i\in H$ only if $p|i$. Thus, $H$, $gH,\ldots,g^{p-1}H$ are pairwise distinct cosets (since $aH=bH$ if and only if $a^{-1}b\in H$), and hence their union equals $G$.
On the other hand, if $G$ is infinite, then since $[G:H]=p$ is finite, we know that there exists a normal subgroup $N$ of $G$ such that $N\subseteq H$ and $[G:N]$ is finite (in fact, divides $p!$). Passing to the quotient $G/N$ and $H/N$ we can reduce the problem to the prior case.