Suppose H be a normal subgroup of G Let H be a normal subgroup of G of index 4  show that there are either exactly 3 or exactly 5 subgroups of G containing  H( including G and H themselves )
Where  do we start from i am completly got no idea..
 A: Hint: $G/H$ is a group of order $4$. And there are only two of them up to isomorphism: $C_4$ or $C_2 \times C_2$. The first has $3$ subgroups, the second $5$. So this boils down to classifying all groups of order $4$. Whether or not you can do that, depends on your group theory knowledge sofar.
A: Use the fourth isomorphism theorem. The groups containing $H$ are in correspondence with the subgroups of $G/H$.
Hence the number subgroups containing $H$ is equal to the number of subgroups of a group $K$ of order $4$. there are only two such groups.
The first is $\mathbb Z_4$ which has $3$ subgroups (the number of subgroups of of a cyclic group of order $n$ is equal to the number of divisors of $n$).
The second is $\mathbb Z_2 \times \mathbb Z_2$ and it has $5$ subgrops ( the number of subgroups of elementary abelian groups with prime $q$ is equal to the number of subspaces when the group is seen as a vector space over $\mathbb Z_q$, these can be calculated in general by using the $q$-binomial coefficients)

In the case in which the elementary abelian group is just $\mathbb Z_q\times \mathbb Z_q$ (dimension $2$) we get $\binom{2}{0}_q+\binom{2}{1}_q+\binom{2}{2}_q= 1+\frac{1-q^2}{1-q}+\frac{(1-q^2)(1-q)}{(1-q)(1-q^2)}=1+(1+q)+1=3+q$

We can solve this problem in general for $p^2$ using the aforementioned tools and these results:


*

*Every group of order $p^2$ is abelian (you prove this by first proving that in any group, if $G/Z(G)$ is cyclic then $G=Z(G)$.

*An abelian group of order $p^2$ is $\mathbb Z_{p^2}$ or $\mathbb Z_p\times \mathbb Z_p$ ( to prove this you could just invoke the fundamental theorem on finite abelian groups or use the well known lemma that if $H$ and $K$ are normal subgroups of $G$ with trivial intersection then $G\cong H\times K$)
Putting this together along with the previous results you obtain that the result must be $3$ or $p+3$. (so in your case $3$ or $5$ because $p=2$)
