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 ?

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Hint: It is enough to show that $G/N$ has a subgroup of index 2 (and thus of order 2). –  Tobias Kildetoft Nov 17 '12 at 20:50

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,$[G:\pi^{-1}(K)] = 2$. - I dont get the last sentence - Hence (if there is a subgroup of order 2 in G/N) there must exist a subgroup of index 2 in G whose image ( what image ? ) is K. Can you explain that please. – André Nov 17 '12 at 23:05 Ok i get this, but how can u prove that this is true, i.e. that if$G/N$has a subgroup of order 2 than$G$has a subgroup of index 2 ? – André Nov 17 '12 at 23:32 Yep but i am just no sure about the last statement :D – André Nov 17 '12 at 23:34 Ok. Sorry but i seem to be stupid :D I hope you mean the canonical projection$\pi: G \rightarrow G/N: g \mapsto gN$. So we got a subgroup$H$of order 2 in$G/N$. We want to show that then there must be a subgroup of index 2 in$G$. Is this subgroup which we are looking for$\pi^{-1}(H)$? And if yes, why has it index 2 ? – André Nov 17 '12 at 23:49 Nice one for the OP.+1 – Babak S. Jan 11 '13 at 19:11 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$G$. -$\mathbb Z /(4)\$ means the modulo 4 group ? –  André Nov 17 '12 at 21:05