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I actually draw the question from (2) of this problem.

I quote it here:

Let $N$ be a normal subgroup of a group $G$ of index $4$. Show

(1) that $G$ contains a subgroup of index $2$,

(2) that if $G/N$ is not cyclic, then there exists three proper normal subgroups $A,B$ and $C$ of $G$ such that $G=A \cup B \cup C$.

When $G/N$ is not cyclic, it must be Klein four-group. I know that then $G/N$ can be a union of three proper subgroups of $G/N$; however, the hint also implies that “thereby $G$ is definitely union of three proper subgroups of $G$”. I can pretty accept it, but feel quite confused when trying to sort it out.

1) How to make it clear to understand? In both senses of strict proof and reasonable thinking.

2) Furthermore, I feel very strongly that quotient groups are just like “contracted groups having the same shape with the original”. But it’s just my imagination, where can I get specific instructions?

Any help is sincerely appreciated.

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Both of these are direct consequences of the Correspondence theorem (or Lattice Isomorphism theorem). It says that there is an order preserving bijection between the sets $$ \{\mbox{Subgroups of }G/N\}\leftrightarrow\{\mbox{Subgroups of }G\mbox{ containing }N\} $$ Moreover, under this isomorphism, normal subgroups correspond to normal subgroups.

1) Since $G/N$ has order $4$, it is isomorphic to $\mathbb{Z}/4\mathbb{Z}$ or $V_4$. In both cases, $G/N$ contains a normal subgroup of index 2, which corresponds to a normal subgroup of index 2 in $G$ (containing $N$).

2) If $G/N\cong V_4$, the three subgroups of index 2 correspond to the 3 subgroups of index 2 in $G$.

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    $\begingroup$ The bijection has even more properties: if $H$ is a normal subgroup of $G$ containing $N$, then $G/H\cong(G/N)/(H/N)$ (sometimes called the third isomorphism theorem, which you actually are using for the answer). $\endgroup$ – egreg Feb 15 '18 at 17:11
  • $\begingroup$ @egreg where is the third isomorphism theorem being used in this context? $\endgroup$ – David Hill Feb 15 '18 at 17:14
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    $\begingroup$ Otherwise, how could you say that $G$ contains a subgroup of index $2$ without using the isomorphism? $\endgroup$ – egreg Feb 15 '18 at 17:16
  • $\begingroup$ fair point. ${\;}$ $\endgroup$ – David Hill Feb 15 '18 at 17:22
  • $\begingroup$ @egreg The third and the forth isomorphism theorem are not the same, here the forth theorem is proper and already sufficient. You can have a look at the solution of this problem and the forth isomorphism theorem(which is also called Correspondence theorem or Lattice isomorphism theorem). Can you see it? $\endgroup$ – Math Feb 16 '18 at 0:36

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