Suppose that $G$ is a group of order $60$ and $G$ has a normal subgroup $N$ of order $2$. Show that $G$ has normal subgroups of order $6$ and $10$
no of $3$-Sylow subgroups are $1$ ,$4$ or $10$ (by Sylow's third theorem)
no of $5$-Sylow subgroups are $1$ or $6$.
If I can prove no of $3$-Sylow subgroups is $1$ and no of $5$-Sylow subgroups is $1$ then both would be normal and I can multiply it with $N$ to get normal subgroups of order $6$ and $10$.
I am trying to use a cardinality argument to prove the same but I am stuck.
fact$1$: Any group of order $30$ order has normal subgroup of order $15$.
proof: If it has a normal subgroup $H$ of order $5$ and $K$ be any sylow-$3$ subgroup of $G$ then $HK$ is a subgroup of $G$ with order $15$ and since $[G:HK]=2$ then $HK$ is normal in $G$. So assume that it has no normal subgroup of order $5$ then $n_5=6\implies$ $6*(5-1)=24$ elements of order $5$. This must require that $n_3=1$ (otherwise you have more than $30$ elements in total.) So, $HK$ is again a subgroup ...
fact$2:$ Any group of order $15$ is cyclic. (you can show it in many ways including Sylow Theorems.)
Now, Let $G$ be a group of order $60$ and $N$ be a normal subgroup of order $2$.
$G/N$ is a group of order $30$ by fact$1$ it has a normal subgroup $H/N$ of order $15$ so $H$ is a normal subgroup of $G$ with order $30$.
Since $|H|=30$ it has a normal subgroup $L$ of order $15$ and $L$ is cyclic and $H$ includes $N$, since $L\cap N=1$, $H\cong L\times N\cong Z_{15}\times Z_2\cong Z_{30}$.
conclusion: Notice that $H$ includes all Sylow-$5$ subgroups since $H$ is normal in $G$ and $5$ divides $|H|$. With same reasoning, it includes all sylow-$3$ subgroups of $G$ . As $H$ has uniqe sylow $p$ subgroups we are done. From that point you can also see that $G\cong Z_{60}$ or $G\cong D_{60}$ or $G\cong Z_2\times Z_{30}$.