Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

Hint: 4th (Lattice) Isomorphism theorem.

share|cite|improve this answer
I have always known this as the `correspondence theorem'. – user1729 Jun 28 '11 at 10:17
Wikipedia ( 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

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

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.