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Find subgroups of a group for a given order

buckeye1973

New member
Feb 22, 2012
5
Hi all,

I'm looking for basic strategies for identifying the subgroups of a group. I believe I have to use conjugacy classes and cycle types, but I'm not sure how to apply those concepts.

Let me pose a specific problem:

Let $G$ be a subgroup of the symmetric group $S_5$, with $|G| = 4$.

By looking this up on a chart, I see that there are three subgroups of $S_5$ of order 4, two Klein-4 groups and the cyclic group $C_4$, so $G$ must be isomorphic to one of these.

How would I go about showing that $G$ must be isomorphic to one of these without looking it up? I'm hoping for a general algorithm here, if possible, such that I could also find subgroups of a certain order of given dihedral groups, for example.

Thanks!

Brian
 

Swlabr

New member
Feb 21, 2012
27
Hi all,

I'm looking for basic strategies for identifying the subgroups of a group. I believe I have to use conjugacy classes and cycle types, but I'm not sure how to apply those concepts.

Let me pose a specific problem:

Let $G$ be a subgroup of the symmetric group $S_5$, with $|G| = 4$.

By looking this up on a chart, I see that there are three subgroups of $S_5$ of order 4, two Klein-4 groups and the cyclic group $C_4$, so $G$ must be isomorphic to one of these.

How would I go about showing that $G$ must be isomorphic to one of these without looking it up? I'm hoping for a general algorithm here, if possible, such that I could also find subgroups of a certain order of given dihedral groups, for example.

Thanks!

Brian
Even for permutation groups, this is not exactly easy. I mean, for a finite group you find all subgroups in finite time by simply taking each subset of your group in turn and checking it is a subgroup...but this isn't particularly fast.

So, to your question here,

If a permutation has order $4$ it is necessarily of the form $(abcd)$ when written in disjoint cycle notation, while if a permutation has order $2$ in $S_5$ it is necessarily of the form $(ab)$ or $(ab)(cd)$ (more generally, it will have the form $(ab)(cd)\ldots (ef)$, but we are working in $S_5$ where such "long" permutations will not have disjoint cycles).

Can you see why these are?

Now, a cyclic group of order $4$ is generated by an element of order $4$, so we have classified all the cyclic groups we had to find (I believe there are $12$ of them - there are $24$ cycles of length $4$, and a $4$-cycle squared is not a $4$-cycle (why?) but the inverse of a $4$-cycle is a $4$-cycle (why?) so we have $\frac{24}{2}$ cyclic subgroups of order $4$). We therefore just need to find the Klein-$4$ groups. These are generated by elements of order $2$, and again we know what they look like. Note also that these elements of order $2$ must commute. That is, if $\sigma_1$ and $\sigma_2$ are your two generators then $\sigma_1\sigma_2=\sigma_2\sigma_1$. This is because the Klein-$4$ group is commutative.

Now, telling you what these groups are is, well, boring, and will teach you nothing. You will learn more if you try and put what I have just told you to the test, and find these groups yourself. I will, however, give you two groups which are isomorphic to the Klein-$4$ group, and two elements which do not generate the Klein-$4$ group. So,

- $(12)$ and $(34)$ generated the Klein-$4$ group.

- $(12)(34)$ and $(13)(24)$ generate the Klein-$4$ group.

- $(12)$ and $(13)$ do note generate the Klein-$4$ group.
 

buckeye1973

New member
Feb 22, 2012
5
OK, thank you. I was thinking along those lines, but had imagined there must be a nicer way. It is good to see it spelled out by someone, so now I have some more confidence that I'm doing things right.

I think I have a strategy at least for keeping it all organized, as follows:

1. List all the possible cycle structures, since all permutations which have the same cycle structure have the same order. E.g., $(123)(45)$ has the same order as $(125)(34)$ and $(12)(345)$, but not as $(12)(34)$.

So I have:
a. $\{1,1,1,1,1\}$ (corresponding to the identity only)
b. $\{2,1,1,1\}$ (e.g., $(12)$)
c. $\{2,2,1\}$ (e.g., $(12)(34)$)
d. $\{3,1,1\}$
e. $\{3,2\}$
f. $\{4,1\}$
g. $\{5\}$ (e.g., $(12345)$)

2. Pick a representative of each cycle structure, and find it's order, so as to know the order of all elements of each cycle structure.

3. By Lagrange's theorem, only elements of order which divide the order of the subgroup in question (in my case 4) can possibly be part of the subgroup. So in my case, those are elements of order 1, 2, or 4. (Which come from a, b, c, and f above.)

4. Select representatives from the conjugacy classes whose elements are order 1, 2, or 4, and try using those to construct groups, and then check the order of those constructed groups.

5. This gives us some obvious choices, such as the cyclic groups generated by elements with cycle structure $\{4,1\}$. I suppose it is just experience that leads us to construct the Klein-4 groups. (Which I have found the two patterns that will produce those.)

6. From there, exhaust the remaining (disjoint) possibilities and demonstrate that they are either not groups or not order 4.

I think that is about it. I will work through the remaining details later, and then maybe try this again with a dihedral group.

Thanks again,

Brian
 

Swlabr

New member
Feb 21, 2012
27
5. This gives us some obvious choices, such as the cyclic groups generated by elements with cycle structure $\{4,1\}$. I suppose it is just experience that leads us to construct the Klein-4 groups. (Which I have found the two patterns that will produce those.)
Yes, experience help a lot. And question such as this one help you build up that experience!

I am tutoring an algebra course at the moment, and the students were asked to write down the subgroup lattice of $S_4$. A hideous, obscenely long question, but one which would have helped the understanding of any student who had attempted it (as it happened, none did!).
 

buckeye1973

New member
Feb 22, 2012
5
I had an exam where we were asked to write the lattice for $D_4$, without ever really being shown how to go about doing so. I worked it out since then, but it took a long time and I never understood how we were supposed to be able to do it on the exam. (Other students agreed, I don't know anyone who got it at the time.) It is the type of thing that many people won't understand until they have seen someone else do it.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
it helps to know that there are just 2 "types" of groups of order 4, the cyclic ones, and the klein group type ones. an easy way to distinguish these particular types is by looking at orders of the elements:

cyclic: elements have order 1,4,2,4
klein 4-group: elements have order 1,2,2,2

only 4-cycles have order 4, so if you find all the 4-cycles, you've found all the possible generators of a cyclic group of order 4. basically there are 5 different ways to pick a 4-element subset of {1,2,3,4,5}, and each one of those ways leads to the same number of 4-cycles that $S_4$ has, which is 6. but if x is an element of order 4, so is $x^{-1}$, so we only get 3 different cyclic subgroups of order 4 for each element of {1,2,3,4,5} that we "leave out", for a total of 15 distinct cyclic subgroups of order 4.

that leaves subgroups isomorphic to the klein 4-group. hopefully you can see that {1,(a b)(c d), (a c)(b d), (a d)(b c)} is always such a group (that gives us 5 possibilities right there, just choose a,b,c and d). but there is another way to get the klein 4-group as well:

<(a b),(c d)> (since disjoint transpositions commute). there are 10 possible choices for (a b), but having chosen (a b) only 3 of the remaining transpositions are disjoint to (a b), which seems like it would mean 30 possibilities...but: <(a b), (c d)> = <(c d), (a b)>, which means we only have 15 subgroups of this type.

that's 35 distinct subgroups of $S_5$ of order 4. i believe that's all of them.

in general, it's not an easy task to determine the subgroup lattice of a given group. the converse of lagrange's theorem, for example, is not true: $A_4$ has order 12, and 6 divides 12, but $A_4$ has no subgroup of order 6.

with permutation groups, we get a little help:

if $\pi = (a_1 a_2 \dots a_{k_1})(b_1 b_2 \dots b_{k_2})\dots(t_1 t_2 \dots t_{k_r})$ where all cycles are disjoint, then:

$|\pi| = \text{lcm}(k_1,k_2,\dots,k_r)$