Prove that there is a subgroup of index 2 Let $G$ be a group, and $N \lhd G$ so that $[G:N]=2014$.
I need to prove that there is a subgroup of index 2.
Indeed, since $N \lhd G$, $o(G/N)=2014$, but I  couldn't find how to go on.
 A: Work in $G/N$:
We have $2014=2\times 19\times 53$.
Hence there exist a subgroup of order $53$. This subgroup is either normal, or there would be at least $54$ such subgroup. Since $53$ is a prime, any 2 subgroup of order $53$ must intersect only at $e$. So there cannot be $54$ subgroup of order $53$. Thus the subgroup of order $53$ is a normal subgroup $H$. Hence we quotient that one out again. And the rest is easy.
Then the rest is just standard correspondence theorem.
A: Hint: use the Sylow-Theorems to show that the group $G/N$ of order $2014=2\cdot 19 \cdot 53$ has a subgroup $H/N$ of index $2$. Then use the Correspondence Theorem to conclude that this corresponds to $H<G$ with $(G:H)=2$.
A: Obviously, we can just replace $G$ with $G/N$, and assume that $G$ itself has order 2014.
We have $2014 = 2 \times 19 \times 53$.
The number of $53$-Sylow subgroups of $G$ divides $2 \times 19 = 38$, but at the same time is congruent to $1 \mod 53$. So there is only one subgroup of $G$ of order 53. Therefore that subgroup is normal. (Any conjugate of it can only be itself.)
So modding out by that subgroup, we can now assume that $G$ has order 38. A $19$-Sylow subgroup will do the trick. 
