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Generally speaking, finding all the subgroups of a finite group is a very difficult problem. But because $S_3$ is finite and of small order, it is possible to use brute force to find all of the subgroups. I need to know both a general method for calculating the subgroup structure by brute force and see how this is applied to the group $S_3$ step-by-step. If I don't see it done step-by-step, I cannot understand it.

Shaun
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Fomalhaut
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1 Answers1

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Here's a brute force method (for finite groups $G$).

First, find all the cyclic subgroups, that is, for each $g$ in $G$, find the subgroup generated by $g$.

Then, find all the two-generator subgroups. For each cyclic subgroup $H$, and each element $g$ in $G$ but not in $H$, find the subgroup generated by $H$ and $g$, that is, the smallest subgroup containing both $H$ and $g$.

Then, find all the three-generator subgroups, then all the four-generator subgroups, etc.

There are shortcuts. Keep in mind that the order of a proper subgroup can't exceed half the order of the group, so as soon as you see that some set of generators gives you more than half the elements of the group, you can stop.

Gerry Myerson
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  • When you say "find all the three-gen subgroups," do you mean take all the subgroups generated by H, g, and find what results when you generate it with h? – Fomalhaut Oct 15 '19 at 02:39
  • I mean for each two-generator subgroup $K$, and each element $s$ of $G$ not in $K$, find the smallest subgroup containing $K$ and $s$. – Gerry Myerson Oct 15 '19 at 02:41
  • See also https://math.stackexchange.com/questions/2082441/how-do-i-find-all-all-the-subgroups-of-a-group – Gerry Myerson Oct 15 '19 at 02:45
  • I was applying your algorithm to $D_8$. I generated the seven groups generated by one element. But because there are seven one-gen groups, and eight elements, there are 7x8 = 56 groups I have to make. Is there a way to reduce this? – Fomalhaut Oct 15 '19 at 08:45
  • Sure. For each one-generator group $H$, you only have to check possible second generators that aren't in $H$. For the one-generator group with four elements, you don't have to do anything – there can't be a proper subgroup with more than four elements. For a two-element $H$, you don't have to check an element of order $4$ as a possible 2nd generator, since that gets you back to something containing a four-element group. (continued) – Gerry Myerson Oct 15 '19 at 08:54
  • If $g$ and $h$ have order two, and you've tried starting with the group generated by $g$, and introducing $h$ to it, then you don't have to check what happens if you start with the group generated by $h$ and introduce $g$ to it. You'll find there aren't really all that many things that need checking. – Gerry Myerson Oct 15 '19 at 08:55
  • So, are we OK now, Balanced? – Gerry Myerson Oct 16 '19 at 22:04
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    Thank you for double-checking on me. I'm not okay because I need to attempt a write-up of an example of me working this out so you can peruse it. I will get one later today. As a professor emeritus you should be used to college students sending you writeups right before midnight ;) – Fomalhaut Oct 16 '19 at 22:07
  • As a retiree, I no longer have to deal with college students sending me writeups, at all. – Gerry Myerson Oct 16 '19 at 23:07
  • If the group is not abelian, then the LCM isn't good enough. E.g., $(1\ 2)$ and $(1\ 3)$ have order $2$ in $S_3$, but they generate all of $S_3$. If $a,b$ both have order $2$, the elements $a,ab,aba,abab,ababa,\dots$ can still all be distinct. It's not a question of enough elements to form a subgroup, it's a matter of the right elements. You just check all products of the elements you have generated to see whether they form a subgroup, or whether you still have to include more elements. But you keep in mind that the order of a subgroup must divide the order of the group. – Gerry Myerson Nov 22 '19 at 01:28
  • I did find an algorithm but I'll share it with you along with its proof of correctness sometime after Dec 2 but before 2020. – Fomalhaut Nov 22 '19 at 03:44
  • I am working on $S_4$. For the first step (all subgroups generated by one element) I had to go over 24 generators. For the second step (all subgroups generated by two elements), I had to go over 24(24 - 1)/2 elements. What ways are there to reduce the number of elements I need to go over with two generators using the subgroups of one generator I found? – Fomalhaut Nov 22 '19 at 07:49
  • Note that if the group generated by $g$ is ${,1,g,g^2,}$, then that's also the group generated by $g^2$; and if the group generated by $g$ is ${,1,g,g^2,g^3,}$, then that's also the group generated by $g^3$, and the group generated by $g^2$ is just ${,1,g^2,}$. So you don't really have to go over 24 generators. I think you get $17$ cyclic subgroups. Then you don't go back to look at all pairs of generators, you look at what happens to each of the cyclic subgroups if you throw in another generator, one that isn't in cyclic group. It's still a lot of work, but (continued... – Gerry Myerson Nov 22 '19 at 08:56
  • ...) some groups have dozens of subgroups, so there's no way around doing lots of work. Still, there are shortcuts. Once you see that $(1\ 2),(1\ 2\ 3)$ generate the symmetric group on the set ${,1,2,3,}$, you know what $(1\ 2),(1\ 2\ 4)$ generate, and more generally what $(a\ b),(a\ b\ c)$ generate. But the way to get a feeling for it is to do it, actually do it for $S_4$. And after you've done it, look back over it and notice all the patterns you missed that could have helped you cut down on the workload. That's how you learn! – Gerry Myerson Nov 22 '19 at 09:00