Finding a normal subgoup Let $G$ be a group with $[G:N] = 4$ where $N$ is a normal subgroup of $G$. I want to show there exists then a subgroup of $G$ with index 2.
How can i approach this problem ?
 A: We know that $|G/N| = 4$, hence $G/N \cong \mathbb Z/(4)$ or $G/N \cong (\mathbb Z/(2))^2$. In either case, $G/N$ has a subgroup of order 2. Its preimage is of index $2$ in $G$.
A: Let $G$ be a group with $[G:N] = 4$ where $N$ is a normal subgroup of $G$.
(Approach?: Definitions are your friend: You are given $N$ is a normal subgroup of $G$, and that its index in $G$ is 4. What does that tell you? How does that related to the quotient group $G/N$? e.g.)
$[G:N]= 4 = |G/N|$. 
Since there are only two groups of order $4$, up to isomorphism, either $G/N \cong \mathbb{Z}_4$ or else $G/N \cong \mathbb{Z}_2^2$. 
In either case, $G/N$ has a subgroup, say $K$, such that $|K| = 2$. (Take, e.g., $K = \{0,2\} \subset \mathbb{Z}_4$ or $K = $K = {(0, 0), (1, 0)}\subset \mathbb{Z}_2^2$.)  
Hence there must be exist a subgroup of $G$ of index $2$ whose image is $K$.

Added for clarification:
For a normal subgroup $N$ of $G$, the group $G/N$ is the quotient group of $G$ by $N$. Recall that the cosets of $N$ in $G$ (since we are given that $N$ is a normal subgroup of $G$) form a group $G/N$ under the binary operation $(aN)(bN)= abN$, where $a, b \in G$. Under this "operation" on $G$ by $N$, so to speak, we have the group $G/N$. Recall the Fundamental Homomorphism Theorem: Every quotient group $G/N$ gives rise to a homomorphism mapping, let's call it $\pi$, $G$ into $G/N$. 
What do you know about the properties of a homomorphism? In this case: If $G/N$ has a subgroup of order $2$, then $G$ must have a subgroup of index $2$. Since we have shown that $G/N$ has a subgroup of order $2$, it follows that $G$ has a subgroup of index $2$.
Recall that $|G/N| = 4$. Every element in $G$ is mapped, via, the homomorphism $\pi$, to an element in $G/N$. There are four such elements in $G/N$, two of which comprise the subgroup $K\leq G/N$. Hence, $\frac{|G|}{|\pi^{-1}(K)|} = 2$. That is, $[G:\pi^{-1}(K)] = 2$.
