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

  1. Look at the order of the group. For example, if it is $15$, the subgroups can only be of order $1,3,5,15$.
  2. Then find the cyclic groups.
  3. 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.

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up vote 7 down vote accepted

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.

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Thanks, that is what i had been wondering. So, let us say D4={e,p1,p2,p3,s1,s2,u1,u2} and {e,u1},{e,u2},{e,p2},{e,s1},{e,s2},{e,p1,p2,p3} are cyclic subgroups. Since there are 5 subgroups of order 2, i have to try all 3 combinations, which is 5 choose 3 =10 possibilities, right? – bigO Mar 16 '13 at 16:17
@bigO you can save some work by noting that if you multiply together two reflections (which are $s_1,s_2,u_1,u_2$ in your notation, and have all order $2$) you will get a rotation. Now there is only one rotation of order $2$ (which should be $p_2$ in your notation), so you don't have to try all possibilities! – Andreas Caranti Mar 16 '13 at 16:22
Thanks, but how do we know that if we multiply two reflections we will get a rotation? – bigO Mar 16 '13 at 18:17
@bigO, the rotations form a subgroup $N$ of index $2$ of $G = D_4$, and the reflections are the other coset, that is, if $r$ is any reflection, then $r N$ is the set of reflections. Now if $r, s$ are two reflections, then since $G/N$ has order $2$, we have $N = rN \cdot sN = rs N$, so the product $rs$ is in $N$. Informally, think of the square with the up side painted white, and the down side painted black. A rotation will leave the white side up, while a reflection will exchange the two sides. If you compose two reflections, you end up with the white side up, so you have done a rotation. – Andreas Caranti Mar 16 '13 at 22:10

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.

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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.

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And what does that have to do with the question? – Tara B Mar 16 '13 at 16:03
Well, OK, it does have a lot to do with the first example, but the question was more general (there was also the example of $D_4$). So I think this is more of a comment than an answer. Then again, so is mine really. – Tara B Mar 16 '13 at 16:05
@TaraB Given the observation 1), this means that the subgroups of order $3$ and $5$ are cyclic. This only leaves the subgroups of $G$ of order $15$... How many are there? – 1015 Mar 16 '13 at 16:06
@julien: You killed the poor $D_4$ at last. Nice + – Babak S. Mar 16 '13 at 16:25
@BabakS. I had no choice, I was forced to! Thanks. – 1015 Mar 16 '13 at 16:31

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.

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What do you mean by see if they generate something you dont already have? So, for example if we think about D4, I have 5 cyclic subgroups of order 2, and 1 cyclic subgroup of order 4. Let them be {e,u1},{e,u2},{e,p2},{e,s1},{e,s2},{e,p1,p2,p3}. And let D4={e,p1,p2,p3,s1,s2,u1,u2}. What should i do after this step? – bigO Mar 16 '13 at 16:10
Well, you don't so much want to 'let' them be some abstract set of letters as to see which elements they actually contain. Then you can see what you get when you multiply elements from different subgroups together. – Tara B Mar 16 '13 at 16:11

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.

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