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

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!!!

• $S_2$ is cyclic of order two. So any element of $S_4$ of order two generates a subgroup isomorphic to $S_2$. So you need to find $9$ elements of $S_4$ of order 2. You have found the two-cycles (of which there are six); are there others? – Chris Eagle Oct 18 '11 at 20:54
• Oh! Are they(13)(24), (12)(34), (14)(23)? – Scharfschütze Oct 18 '11 at 21:07
• Yes.${}{}{}{}{}$ – Chris Eagle Oct 18 '11 at 21:07
• @Scharfschütze: Your method not only found group isomorphisms, but permutation group isomorphisms (it worked by relabeling the points, not just relabeling the group elements). Unfortunately, your book used "S2" to mean the cyclic group of order 2, not the permutation group acting non-trivially on 2 points, so you get those three extra groups that act on 4 points. – Jack Schmidt Oct 18 '11 at 21:14
• @Scharfschütze: May I suggest posting your solution as an "answer"? That way the question won't go unanswered, even though you already know the answer. – Arturo Magidin Oct 18 '11 at 22:14

## 1 Answer

Here is a recipe for finding all subgroups of Sn 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 Sn. 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.

• (I leave out the normalizer calculations that Scharfschütze already seems to understand; since they are permutation actions, one is just looking for ways of relabeling the points without changing the group.) – Jack Schmidt Oct 18 '11 at 23:04
• math.stackexchange.com/questions/66543/… asks if there is a similar method when the big group is a projective general linear group. – Jack Schmidt Oct 19 '11 at 0:59