Proving that there exists a subgroup of $G/N$ that's isomorphic to $H\le G,\ \text{with} \ |H|=p$ and $H\nleq N$ Let $G$ be a group with a normal subgroup $N$ and a subgroup $H$ with $|H|=p$ of prime cardinality and $H$ is not a subgroup of $N$.
The question is to prove that $G/N$ has a subgroup that is isomorphic to $H$.
My attempt:
$|H|=p\implies $ $H$ is cyclic. $H=\langle h\rangle$ for some $h\in H$
The correspondance theorem states that $f:\{K\le G\ :\ K\supset N\}\rightarrow\{\text{subgroups of } G/N\}$ is a bijection but this doesn't seem to be useful here.
A hint would be welcome
 A: Since $H$ is not contained in $N$, there exists a nonidentity element $g\in H\setminus N$. It follows that $gN$ is not identity in $G/N$. Because the order of $g$ is $p$, $(gN)^p=1$ in $G/N$. Hence, the order of $gN$ in $G/N$ divides $p$, and the order must be $p$. Now, the subgroup of $G/N$ generated by $gN$ has order $p$, which is surely isomorphic to $H$.
A: The point is that since $H$ is not a subgroup of $N$, in fact $H \cap N = \{e\}$, since if $h^r \in N$ for some $r$, then unless $r  \equiv 0 \mod p$ it has a multiplicative inverse modulo $p$ say $rr' \equiv 1 \mod p$, in which case $(h^r)^{r'} = h \in N$, so $H$ will be a subgroup of $N$,a contradiction. 
Now, we may proceed in this way : consider the cyclic group $\langle hN\rangle \subset G/N$. We note that it has at most $p$ elements since $h^pN = N$ as $h^p = e$. However, if $h^rN =N$ for some smaller $r$, then this implies that $h^r \in N$, which is a contradiction as we have $h^r \in H$, and showed earlier that $H \cap N = \{e\}$. So, the cyclic subgroup $\langle hN\rangle$ of $G /N$ is isomorphic to $H$, since any two groups of the same prime order are isomorphic.
Alternately, use the second isomorphism theorem : note that $N$ is normal in $G$, and therefore $NH=HN$ is a subgroup of $G$. The second isomorphism theorem asserts that $NH/N \cong H /(H \cap N)$, but we have already seen that $H \cap N = \{e\}$, so $HN/N \cong H$. So one deduces that $HN/N$ is a subgroup of $G/N$ which is isomorphic to $H$.
A: Hint:
$N=${$x^p|\forall x\in G$}
Step 1: Show N is normal subgroup of G and does not contain element of order p .
Step 2:
Define  $\phi: H\to G/N$
H is cyclic group 
say H$=<a>$
$\phi(a^i)=a^iN$
Show map is well defined, Bijective
A: Hint:Let $H=\langle h\rangle$ for some $h\in H$. Define $P\subseteq G/N$ such that  $P=\{h^{m}N \mid m\in \Bbb{Z}\}$.First prove that $P$ is a subgroup of $G/N$. After that, define a map $\psi: P \to H$ such that $\psi(h^mN)=h^m$. Then show the map is a well-defined homomorphism and a bijection.  
