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.
A simpler approach, let $P \in Syl_3(G)$ and assume $G$ to be simple. Then $|G:P|=4$. $G$ acts by left multiplication on the left cosets of $P$ and the kernel of this action is core$_G(P)=\bigcap_{g \in G}P^g$. This is a normal subgroup of $G$ and since $G$ is simple, it it trivial. This means that $G$ embeds isomorphically into $S_4$. This implies $36 \mid 24$, by Lagrange's Theorem, a contradiction. Hence $G$ cannot be simple.
Note on the solution The following is a generalization of the well-known Cayley imbedding, which can be recovered by taking $H=1$ in the sequel.
Let $H \leq G$, with $|G:H|=n$ a positive integer. Put $\Omega=\{g_1H,g_2H, \ldots ,g_nH\}$, the set of left cosets of $H$. $G$ works on $\Omega$ by multiplication from the left: if $g \in G$, then $g(g_iH)=gg_iH$ is again one of the elements of $\Omega$. So each $g \in G$ induces a permutation $\sigma_g$ on $\Omega$. Hence we can define a map $f: G \rightarrow S_n$, by $f(g)=\sigma_g$. It is easy to verify that $f$ is a homomorphism (observe that $\sigma_g \circ \sigma_h=\sigma_{gh}$ for all $g,h \in G$). Now let us compute ker$(f)$. If $x \in $ ker$(f)$, this means that $x$ induces the trivial permutation, so for any left coset $gH$, we must have $xgH=gH$, implying $x \in gHg^{-1}$ for all $g \in G$. This amounts to $x \in \underset{g \in G}\bigcap H^g$ (where $H^g=g^{-1}Hg$; note that when $g$ runs through the entire $G$, $g^{-1}$ does the same, so intersection over $g^{-1}Hg$ is the same as over $gHg^{-1}$). Conversely, if $x \in \underset{g \in G}\bigcap H^g$, then it induces a trivial permutation.
We conclude that ker$(f)=\underset{g \in G}\bigcap H^g$ and by the first isomorphism theorem $G/$ker$(f) \cong $ im$(f)$, being a subgroup of $S_n$.
In general, if $H \leq G$, the core$_G(H)$ is the largest normal subgroup of $G$ contained in $H$. It is easy to show that core$_G(H)=\underset{g \in G}\bigcap H^g$.
Best Answer
To simplify notation, let me write $S_{12}=S_1\cap S_2$, $S_{123}=S_1\cap S_2\cap S_3$, etc. Define $f:S_{12}/S_{123}\to S_2/S_{23}$ by $f(aS_{123})=aS_{23}$. To prove this is well-defined, we must show that if $aS_{123}=bS_{123}$ for $a,b\in S_{12}$ then $aS_{23}=bS_{23}$. But if $aS_{123}=bS_{123}$ then $b^{-1}a\in S_{123}$ which implies $b^{-1}a\in S_{23}$ so $aS_{23}=bS_{23}$.
Now I claim that $f$ is injective, and thus $[S_{12}:S_{123}]\leq[S_2:S_{23}]$. To prove this, suppose $a,b\in S_{12}$ are such that $aS_{23}=bS_{23}$. This means $b^{-1}a\in S_{23}$. Since $a,b\in S_{12}$, we also have $b^{-1}a\in S_1$, so $b^{-1}a\in S_{123}$. Thus $aS_{123}=bS_{123}$. That is, $f(aS_{123})=f(bS_{123})$ implies $aS_{123}=bS_{123}$, so $f$ is injective.
More generally, the same argument shows that if $A$ is a group with subgroups $B$ and $C$, then $[C:B\cap C]\leq [A:B]$. (In your case, we have $A=S_2$, $B=S_{23}$, and $C=S_{12}$.)