I am new with abstract algebra and I trying to find all the subgroups of $S_{4}$ generated by the cycles :

a) $(13)$ and $(1234)$

b) all cycles of length $3$

I am not sure how to start so I would very much appreciate some hints or at least some material to get started. Thank you

  • $\begingroup$ If you are trying to enumerate the elements of a subgroup generated by some set of elements, you can start by just listing all the products you can in some systematic way. For example, if you were looking for $\langle 2,3 \rangle \leq \mathbb{Z}_{25}$, start by finding out what $2, 3, 2+3, 2 + 3 + 3, 2 + 2 + 3, \dots$ are equal to. Then you can get a feel for what you think the subgroup should look like. Good luck! $\endgroup$ – Bruce Zheng Oct 7 '14 at 20:05

I assume that in each case you're actually looking for answers to a two-part question:

  1. What is the set of elements in the subgroup?
  2. What well-known group is the subgroup isomorphic to?

Answering #1 is relatively easy. Start with the set of elements you've been given. So, in (a), start with the set $\{(13), (1234)\}$, and in (b), start with the set of all cycles of length 3. Then, multiply arbitrary pairs of elements together (multiplying an element with itself is also allowed). In the context of the symmetric group $S_4$, "multiplication" just means function composition.

Consider case (a), and let $x = (13)$ and $y = (1234)$. You'll start by computing all possible pairs of multiplications $xx$, $xy$, $yx$, and $yy$.

Write the results in cycle notation, and determine whether any of those results represent permutations that aren't in your set. If they're new, add them to the set, and continue the process. You may end up computing results such as $xyyxxy$, but eventually you'll run out of "new" results; that is, you'll only find results that simplify to elements that are already in your set. When you can't find any way combine elements in your set to generate a value that lies outside of your set (in other words, as soon as the set becomes "closed" with respect to the group operator), you'll have found the full subgroup.

For #2, you can try to determine whether your group is isomorphic to any well-known group, such as those listed here: http://en.wikipedia.org/wiki/List_of_small_groups

Roughly speaking, "isomorphic to" means "behaves like". So, to answer this part of the question, you can use tricks like the ones listed in other answers to narrow down the possibilities.


$a)$ subgroup generated by $(13) $ and $(1234)$ is isomorphic to $D_4$. Write its elements precisely and then you will at once see it, four rotations of square (including $0$ degree) and four flips.

b) Subgroup of $S_4 $ generated by all $3$ cycles is $A_4$, it is true for any $S_n$, but sorry man, you will have to work out explicitly all elements for $S_4$ (or you won't learn) as there are not many elements and without much time consumption you can write both subgroups you asked for.

I have told you the answer, now verifying is your job

  • $\begingroup$ Thanks for the hints I have managed to work on the problem. $\endgroup$ – John Oct 7 '14 at 21:44

$a)$ Let $H=\langle(1,2,3,4)\rangle$ and $k=(1,3)$ you can show that $kHk^{-1}=H$ which means $H\langle k\rangle$ is agroup of order $8$.

$b)$ Notice that a three cycle is an even permutation so group generated by three cycles must be a subgroup of $A_4$. And this group has elements of order three and two. Since $A_4$ has no subgroup of order $6$, this group is $A_4$.


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