Subgroups of a group of order 60 with a normal subgroup of order 2 (Sylow) This is the problem 38 of the chapter 24 in the Gallian's Algebra. 

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, 10, 30.
  
*$G$ has subgroups of order 12 and 20.
  
*$G$ has a ciclic subgroup of order 30.
  

I've been dealing with this problem, but it seems i don't have good ideas. I could only prove the first item for the order 30.
Can someone give me a hand please? Thanks!
 A: Ok, I've got it. Order 30 is tricky. As in Devilathor's comment, create a subgroup $H$ of $G/N$ with order 15. Then $H$ is normal in $G/N$. However, since $|H| = 15$, it must have a normal Sylow 5-subgroup no matter what. Now the trickier bit - $|H| \cong \mathbb{Z}_{15}$. Since $P_{3}$ normalizes $P_{5}$, it must act on $P_{5}$ by conjugation. However, $P_{3} \cong \mathbb{Z}_{3}$, which can only act trivially on $P_{5}$, since $|\textrm{Aut}(P_{5})| = 4$. So $P_{3}$ and $P_{5}$ commute, and $H \cong \mathbb{Z}_{15}$. (That's as far as we can go with $H$, since it might be $D_{30}$, whose Sylow 2-subgroup is not normal.)
Now, we have normal 3- and 5-subgroups of $H$, which are actually characteristic in $H$, so they're normal in $G/N$ as well. That means when we pull them upstairs to $G$, they're normal in $G$. That gives us our normal subgroups for part 1.
For part 2, we pull a similar trick - the normal subgroup of order 6 has a characteristic Sylow 3-subgroup, so we have $Q_{3}$, a normal Sylow 3-subgroup of $G$ - multiply that by a Sylow 2-subgroup and you get a subgroup of order 12. Then do the same with the normal subgroup of size 10 and $p=5$. 
For part 3, just multiply together $Q_{3}$, $Q_{5}$, and $N$. Those are normal subgroups of $G$ that don't intersect, so they commute and their product is Abelian.
