Subgroups of $S_4$ isomorphic to $S_3$ and $S_2$?

Question

It's a home work problem I got: Find $4$ different subgroups of $S_4$ isomorphic to $S_3$ and $9$ isomorphic to $S_2$.

My approach is: since $S_3=\{1, (123),(132),(12),(23),(13)\}$, just take the groups of permutations on $\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\}$, obviously they are all subgroups of $S_4$.

To find the isomorphism, for each subgroup, just assign 1 to the first element, 2 to the second, 3 to the third. This is going to be an isomorphism.

For example, in the group of permutations on $\{1,2,4\}$, assign $(124)\rightarrow(123) (24)\rightarrow(23),(142)\rightarrow(132), (14)\rightarrow(13)$, etc.

I think this method is fine, but I have trouble with the second part of the problem. That is, to find the $9$ different subgroups of $S_4$ isomorphic to $S_2$. When I pick $2$ elements out of $\{1,2,3,4\}$, there can only be $\frac{4!}{2!2!}=6$ ways, which means this method only gives $6$ different subgroups isomorphic to $S_2$, but the problem says there are $9$.

Is my method wrong? Or are there some other subgroups that I've missed?

Thanks!!!

Answer 1

Here is a recipe for finding all subgroups of S_n that are isomorphic to a fixed and completely understood group G, like G = S₃.

First, find all conjugacy classes of subgroups H of G, and label them by their index $[G:H] = |G| / |H|$. For example, for G = S₃:

• "1" is the subgroup { 1, (123), (132), (12), (13), (23) }
• "2" is the subgroup { 1, (123), (132) }
• "3" is the subgroup { 1, (12) }, or one of its conjugates: { 1, (13) } or { 1, (23) }
• "6" is the subgroup { 1 }

Each of these defines a way for G to act (multiplication on the cosets of H in G). For instance, for G = S₃:

• "1" is the action on 1 point { A }, where every element of G leaves A alone
• "2" is the action on 2 points { B, C }, where the 2-cycles { (12), (13), (23) } all switch B and C, but everyone else { 1, (123), (132) } leaves A and B alone
• "3" is the action on 3 points { 1, 2, 3 }, where everybody does the natural thing
• "6" is the action on 6 points { E = 1, F = (123), G = (132), H = (12), I = (13), J = (23) }, where everybody acts as multiplication

Now write n as an unordered sum of the indices. For instance, if n = 4, then there is only one way:

• "4 = 3 + 1" acts on 1,2,3, and A=4. The combined action is { 1, (123), (132), (12), (13), (23) }, where nobody moves 4.

A better example is n = 5, where there are two ways:

• "5 = 3 + 1 + 1" acts on 1,2,3 and leaves 4 and 5 alone: { 1, (123), (132), (12), (13), (23) }
• "5 = 3 + 2" acts on 1,2,3 and B=4, C=5: { 1, (123), (132), (12)(45), (13)(45), (23)(45) }

Maybe even better is n = 6, where there are four ways:

• "6 = 3 + 1 + 1 +1" is { 1, (123), (132), (12), (13), (23) }
• "6 = 3 + 2 + 1" is { 1, (123), (132), (12)(45), (13)(45), (23)(45) }
• "6 = 3 + 3" needs {4,5,6} to be a second copy of {1,2,3}, and so it is: { 1, (123)(456), (132)(465), (12)(45), (13)(46), (23)(56) }
• "6 = 6" is just { 1, (123)(456), (132)(465), (14)(26)(35), (15)(24)(36), (16)(25)(34) } = { 1, (EFG)(HIJ), (EGF)(HJI), (EH)(FI)(GJ), etc.}

In each case, we are writing a general action of G as a sum of simple "transitive" actions of G on the cosets of a subgroup H. This is called the orbit stabilizer theorem; your course should mention this at least superficially. The idea of literally adding them up is one way to view "permutation characters" in character theory; that will be in a later course, most likely.

This only works if (a) you understand G very well, and (b) the "big" group is S_n. A similar thing works if the "big" group is a general linear group, but then the H are replaced by modules and this is called representation theory of finite groups.