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I've got 2 permutations $x,y \in S_4$. I'm supposed to determine (all the elements of) $\langle x,y \rangle$ (and not just identify this subgroup by saying what it's isomorphic to or whatever). If it makes things any easier, then: I have a transposition $x$ and a 4-cycle $y$. Rather than giving the specific $x$ and $y$, I wanna know how to tackle this kind of problem in general.

With the specific $x$ and $y$ given, I was able to determine that there were at least 12 members in this subgroup.

Method 1 (Lagrange) : Well Lagrange says the number of elements is either 1,2,3,4,6,8,12,24. After finding 12, by Lagrange, all I have to do is find an element of $S_4$ that isn't in the subgroup to prove it's not $S_4$ and thus it must not contain anything other than those 12 elements.

  • Unfortunately, I'm not allowed to use Lagrange...but I guess I can if I can somehow prove Lagrange without cosets (and use only tools before cosets in your basic abstract algebra book i guess) or at least that the subgroups of $S_n$ have orders that divide $n$. But wait even if I did use Lagrange or something similar, I think it's pretty tough to show it doesn't belong. Idk.

Method 2 (multiplication table): After finding 12, perform 144 multiplications (minus the ones I already did in finding the 12 elements). Then see that each of the 144 multiplications result in 1 of the 12 elements.

  • (2.1. This uses Lagrange but only in working backwards! After I do the 144 multiplications, I can just claim luck. It's like proving matrix $A$'s inverse $A^{-1}=B$ by coming up with some $B$ from wolfram and then just showing $BA=I$. I don't have use whatever formula to come up with the inverse manually.)

  • 2.2. Wait actually I don't think this is necessarily helpful...what if there's some 13th (and thus possibly more) element that is indeed part of the group but the 1st 12 elements simply don't cover this 13the element? Hmmm...idk.

  • 2.3. I've got it! A way to list the elements of subgroup generated by certain elements is like do a multiplication table with elements you have (and identity). And then keep repeating with the new elements and old elements and until you get only old elements? Is there some theorem of multiplication tables like this?

Method 3: got anything else?

Related questions:

  1. order of $\langle (123) , (234) \rangle$
  • Wow what a joy Lagrange and orbit-stabiliser theorems are! I have a newfound appreciation for these theorems.
  1. Determine subgroup $\langle (12),(13)(24)\rangle $ of group $ S_{4}$
  • ok now this seems like my kind of answer. There seems to be some trick here. Something about

$$\langle (12),(13)(24)\rangle = \langle (12),(12)(13)(24)\rangle = \langle (12),(3241)\rangle$$

What's going on?

  1. Finding Cyclic Permutation Products
BCLC
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    It seems that you missed a subgroup of order $8$ in $S_4$, which is a Sylow $2$-subgroup and isomorphic to the dihedral group $D_8$. If you know the group structure of each order, then it might help. For example, the only subgroup of order $12$ is $A_4$, and all subgroups of order $8$ is isomorphic to $D_8$. For $6$, it must be $S_3$ because there is no element of order $6$ in $S_4$. It is easier to determine $\langle x, y\rangle$ once you know all the possibilities. – Groups Oct 08 '21 at 20:58
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    Also, any group generated by two elements of order $2$ is dihedral, which directly gives my answer to the question of your second link. – Groups Oct 08 '21 at 21:00
  • @HongyiHuang added 8. thanks! you can post as answer. so if i get 12 elements then how would i show it's everything: prove it's isomorphic to get $A_4$? – BCLC Oct 11 '21 at 15:35
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    To prove the subgroup of order $12$ is $A_4$: see https://math.stackexchange.com/questions/27024/a-n-is-the-only-subgroup-of-s-n-of-index-2 – Groups Oct 11 '21 at 21:09
  • @HongyiHuang wow thanks. you can post that as answer even if it's not necessarily complete. i think practically it should be complete though – BCLC Oct 13 '21 at 12:31

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