How to find non-cyclic subgroups of a group? I am trying to find all of the subgroups of a given group. To do this, I follow the following steps:


*

*Look at the order of the group. For example, if it is $15$, the subgroups can only be of order $1,3,5,15$.

*Then find the cyclic groups.

*Then find the non cyclic groups.


But i do not know how to find the non cyclic groups. For example, let us consider the dihedral group $D_4$, then the subgroups are of the orders $1,2,4$ or $8$. I find all cyclic groups. Then, I saw that there are non-cyclic groups of order $4$. How can I find them? I appreciate any help. Thanks.
 A: In the $n=15=3\cdot 5$ case, recall that every group of order $p$ prime is cyclic. This leaves you with the subgroups of order $15$. How many are there?
Of course, this is not as easy in general. For general finite groups, the classification is a piece of work. Finite Abelian groups are easier, as they fall in the classification of finitely-generated Abelian groups.
Now, $D_4$ is not that bad. The only nontrivial thing is to find all the subgroups of order $4$. Cyclic ones correspond to order $4$ elements in $D_4$. Noncyclic ones are of the form $\{\pm 1,\pm z\}$ where $z$ is an order $2$ element in $D_4$. Since $D_4$ has eight elements, it is fairly easy to determine all these order $2$ and $4$ elements.
A: One thing you can try is find the groups of each order. A group of order $2$ must be isomorphic to $\mathbb{Z}_2$, which contains identity and another element of order $2$. How many elements of order $2$ are there?
For groups of order $4$, they are isomorphic to either $\mathbb{Z}_4$ or $\mathbb{Z}_2\times\mathbb{Z}_2$. In $\mathbb{Z}_4$, it contains an element of order $4$, so what is it? The other case is similar.
You can also find them using Group Explorer.
A: In general, finding the subgroups (of a biggish group) won't be easy.
In the case of $D_4$, the only non-cyclic groups besides $D_4$ itself can only be of order $4$. So you are looking at subgroups of $G$ that consist of the identity, and three involutions (elements of order $2$) $a, b, c = ab$.
Now try out the various possibilities, avoiding repetitions.
A: For any two subgroups $U,V \leq G$ there is a unique smallest subgroup $\langle U,V \rangle$ above $U$ and $V$. It consists of all finite products of elements of $U$ and $V$. You can gain all subgroups of $G$ by starting with the cyclic subgroups and iterating that operation until no new group occurs.
A: You could always start by taking the generators of two of your cyclic subgroups and see if they generate something you don't already have.
A: This statement might help you 
" for every subgroup of dihedral group ,either every member of the subgroup is a rotation or exactly half of the members are rotations "
