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I am stuck with a problem I found in a book on theory of groups. Here it is:

Let $N$ be a normal subgroup of a group $G$ of index 4. Show (1) that $G$ contains a subgroup of index 2. Show (2) that if $G/N$ is not cyclic, then there exists three proper normal subgroups $A,B$ and $C$ of $G$ such that $G=A \cup B \cup C$.

This problem arises in a chapter devoted to the homomorphism and isomorphism theorems.

Any hint?

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3 Answers 3

Hint: 4th (Lattice) Isomorphism theorem.

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I have always known this as the `correspondence theorem'. –  user1729 Jun 28 '11 at 10:17
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Wikipedia (en.wikipedia.org/wiki/Lattice_theorem) gives synonyms for the name of this theorem. –  Thomas Connor Jun 28 '11 at 10:19
    
    
OK, I think I got it. Thanks for your advice. In fact, the 4th isomorphism theorem isn't mentioned in my book. It works perfectly. Would you like to write a complete answer so that I can accept it or do you want to let me do so? –  Thomas Connor Jun 28 '11 at 10:55
up vote 4 down vote accepted

Here is a proof of the results (1) and (2) based on tips given by lhf and jspecter.

We have $\vert G : N \vert = 4$, hence $G/N \cong C_4$ or $C_2 \times C_2$. If $G/N$ is not cyclic, then $G/N \cong C_2 \times C_2$ and (2) is clear from the multiplication table.

Let $\mathcal{G} = \{ A \mid N \leq A \leq G\}$ and les $\mathcal{N}$ be the set of all subgroups of $G/N$. Then the 4th isomorphism theorem states that $$ \phi : \mathcal{G} \to \mathcal{N}: A \mapsto A/N$$ is bijective. Thus $\vert \mathcal{G} \vert = \vert \mathcal{N} \vert = 4$. Whether $G/N \cong C_4$ or $G/N \cong C_2 \times C_2$, it is clear that $G/N$ contains a subgroup $H$ of index 2. Hence, there exists $K \leq G$ such that $\phi(K) = K/N = H$. By virtue of the 4th isomorphism theorem, we have $\vert G/N:K/N\vert = \vert G:K \vert = 2$ and (1) is proven.

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$G/N$ is (isomorphic to) either $C_4$ or $C_2 \times C_2$. What are the subgroups of $C_2 \times C_2$ ?

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