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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

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up vote 3 down vote accepted

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, $[G:\pi^{-1}(K)] = 2$.

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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 –  B. 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 $G$.

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$\mathbb Z /(4)$ means the modulo 4 group ? –  André Nov 17 '12 at 21:05
    
Yes, it does, the cyclic group of order 4. –  martini Nov 17 '12 at 21:07

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