I got stuck on this specific problem:
Let $G$ be a group and $H \mathrel{\unlhd} G$ with $|G : H| = p^k$ with $p$ is a prime and $ k \geq 1$.
Prove that there exists a $g \in G$ such that $g \not\in H, g^2 \not\in H, g^3 \not\in H, ... g^{p-1} \not\in H$ but $g^p \in H$.
So we know that there are $p^k$ cosets and $H$ is a normal subgroup of $G$ ($gH = Hg, \forall g \in G$).
From here, I have no idea how to find that such $g$.
Thanks in advance.
Since
$H \mathrel{\unlhd} G, \tag 1$
the collection of (left or right) cosets of $H$ form a group $G/H$; the order of this group is the index $[G:H]$ of $H$ in $G$; thus
$\vert G/H \vert = [G:H] = p^k, k \ge 1; \tag 2$
thus
$p \mid \vert G/H \vert, \tag 3$
and hence it follows from Cauchy's theorem that $G/H$ contains an element of order $p$; such an element of $G/H$ is then a coset of the form $gH$ for some $g \in G$, and since
$\vert gH \vert = p, \tag 4$
it follows that
$(gH)^k \ne e_{G/H}, \; 1 \le k \le p - 1, \; (gH)^p = e_{G/H}, \tag 5$
where $e_{G/H}$ denotes the identity element of the group $G/H$, that is, the coset $eH$ where $e$ is the identity element of $G$. Now
$(gH)^k = g^kH, \tag 6$
and
$g^kH = e_{G/H} \Longleftrightarrow g^k \in H; \tag 7$
it follows then that
$g^k \notin H, \; 1 \le k \le p - 1, \; g^p \in H, \tag 8$
which was to be proved.