Group $G$ with $N\unlhd G$, $|N|=5$ and $G/N\cong S_3$ Group $G$ with $N\unlhd G$, $|N|=5$ and $G/N\cong S_3$.
Question is : Let $G$ be a group with a normal subgroup $N\unlhd G$ of order $5$, such that $G/N\cong S_3$. Show that, $|G|=30$, $G$ has a normal subgroup of order $15$, and $G$ has $3$ subgroups of order $10$ that are not Normal.
Intuitively, this is very clear But I am Not so sure how to write this in detail.
As $|N|=5$ and $|S_3|=6$ and $G/N\cong S_3$ we have $|G/N|=|S_3|$ which implies $|G|=|N|.|S_3|=5.6=30$ (DONE).
$S_3$ has Normal Subgroup of order $3$ which is $H=\{(1)(2)(3), (123),(132)\}$, and so, I somehow see that $H\times N$ is normal group with $|H\times N|=|H||N|=3.5=15$ (PARTIALLY DONE)
$S_3$ has $3$ Non-Normal (:)) subgroups  of order $2$. $H_1=\{(1)(2)(3), (12)\}$, $H_2=\{(1)(2)(3), (13)\}$, $H_3=\{(1)(2)(3), (23)\}$, We consider $N\times H_1$, $N\times H_2$, $N\times H_3$ are required subgroups of order $10$ which are Non-normal. (PARTIALLY DONE)
I am not very sure about the way i have Justified this. 
Please let me know if there is any possible better way for this.
Thanks.
 A: In fact when $G/N\cong S_3$ so $G$ is an extention of $N$ by $S_3$ so $G\cong N\rtimes S_3$ and this gives you $|G|=30$. Also, according to well-known fact:

If $|G|=2n$ in which $n$ is odd then $G$ has a subgroup of order $n$.

our group $G$ has a subgroup of order $15$ which is also a normal one. If you are familiar to the presentaion of groups, with a good knowledge of semidirect product, then you may find the following presentaion for $G$:
 $$G=\langle a,b\mid a^2=1=abab^4\rangle$$
The following codes might help you when using GAP:
   > f:=FreeGroup("a","b");;
   > a:=f.1;;  b:=f.2;; 
   > s:=f/[a^2,a*b*a*b^4];;
     Order(s);
     StructureDescription(s);
     IsAbelian(s);

     30
     "C5 x S3"
     false

A: This is basically just an application of the lattice isomorphism theorem.
For example, you noted that $S_3$ has a subgroup $H$ of order $3$ which is normal. Then, by the lattice isomorphism theorem there exists a subgroup $N\subseteq K\subseteq G$ such that under the projection $G\to G/N$ one has that $K$ mapsto $H$. You know that this has order $15$ since $K/N$ has order $3$. You also know it's normal since the isomorphism $G\to S_3\to S_3/H$ has kernel $K$. 
Try and apply this logic to solve your other question.
