Every group of order $60$ , having a normal subgroup of order $2$ , has a normal subgroup of order $6$ (without Sylow )? How to prove , without using Sylow's theorems , that every group of order $60$ , having a normal subgroup of order $2$ , contains a normal subgroup of order $6$ ? Please help . Thanks in advance
 A: FACT 1: Group of order $15$ is cyclic. [without Sylow theory]
By Cauchy theorem, $G$ contains a subgroup $H$ of order $5$; if $H_1$ is another subgroup of order $5$ then $|HH_1|=|H|.|H_1|/|H\cap H_1|=5.5/1>|G|$, contradiction. So subgroup of order $5$ is unique, hence normal. Let $K$ be a subgroup of order $3$. 
For every $k\in K$, define $\varphi_k\colon H\rightarrow H$, $\varphi_k(h)=khk^{-1}$, it is an automorphism (since $H$ is normal). We can see that $\varphi_k\varphi_{k'}=\varphi_{kk'}$; hence $k\mapsto \varphi_k$ is a homomorphism from $K$ to $Aut(H)$. 
Since $K\cong \mathbb{Z}_3$ and $Aut(H)\cong \mathbb{Z}_4$, the homomorphism from $\mathbb{Z}_3$ to $\mathbb{Z}_4$ is trivial. This means $\varphi_k=$identity for every $k\in K$; this means, $khk^{-1}=h$ for all $h\in H$ (and $k\in K$). This means $H$ and $K$ commute elementwise. Therefore $G=H\times K=\mathbb{Z}_5\times \mathbb{Z}_3$, which is cyclic.
FACT 2: Any group of order $2m$ ($m$ odd) has normal subgroup of order $m$. 
This is also easy to prove without Sylow; you can try (or then search for link).
Towards your question: Let $N$ be normal subgroup of order $2$. Then $G/N$ is group of order $30=2.15$. Now $G/N$ has normal subgroup, say $L/N$ of order $15$; it should be cyclic; hence it has unique subgroup of order $3$, say $H/N$. So we have situation: 
$H/N$ is unique subgroup of order $3$ in $L/N$ and $L/N\trianglelefteq G/N$. 
Exercise: Show that $H/N$ is normal in $G/N$.
Then $H$ is normal in $G$ with $|H|=|N|.|H/N|=2.3=6$. 
