proof that the alternating group is simple on 5 or more elements, and some interesting consequences
This post has a few objectives: prove that the alternating group \(A_n\) is simple, and then conclude that \(S_n\) is not solvable, both for when \(n \geq 5\). In order to do this, we’ll first discuss some facts about \(2\)-transitive groups, and then see a few examples when \(n < 5\). After this post, I’ll hopefully start to talk about some basic results of Galois Theory (as soon as I figure out a good way of plotting tikzcd diagrams on this blog).
If \(G\) is a group that acts on a set \(X\), i.e., if there is a homomorphism from \(G\) to \(\text{Sym}(X)\), then we say that \(G\) acts transitively on \(X\) if for every \(\alpha,\beta \in X\), there exists some \(g \in G\) such that \(\alpha g = \beta\). In other words, for every element \(\alpha \in G\), its orbit \(\alpha G\) is the entire set \(X\). A group is said to be \(2\)-transitive on \(X\) if, for every \(\alpha,\beta,\gamma,\delta \in X,\alpha \neq \beta, \gamma \neq \delta\), there exists some \(g \in G\) such that
\[(\alpha,\beta)g = (\alpha g,\beta g) = (\gamma,\delta)\]Equivalently, a group is \(2\)-transitive on \(X\) if it is transitive on the set of distinct ordered pairs \(\{(\alpha,\beta)\vert \alpha,\beta \in X,\alpha \neq \beta\}\). From this definition, it follows that any \(2\)-transitive group is also transitive, and we can also prove the following characterization.
Proposition 1: A group \(G\) is \(2\)-transitive on \(X\) if, and only if, the stabilizer \(G_\alpha\) of any element \(\alpha \in X\) acts transitively on \(X\setminus\{\alpha\}\).
Proof: First, assume that \(G\) is \(2\)-transitive, and let \(G_\alpha\) be the stabilizer of some \(\alpha \in G\). By definition, we know that for any \(\beta,\delta \in X\setminus\{\alpha\}\), there exists some \(g \in G\) such that
\[(\alpha,\beta)g = (\alpha,\delta)\]hence \(\beta g = \delta\) and \(\alpha g = \alpha\), and thus \(g \in G_\alpha\), implying that this stabilizer indeed acts transitively on \(X\setminus\{\alpha\}\).
Now if \(\alpha,\beta,\gamma,\delta \in X\), with \(\alpha \neq \beta, \gamma \neq \delta\), we can find elements \(g_1,g_2 \in G\) such that
\[\begin{split} (\alpha,\beta)g_1 &= (\alpha,\delta)\\ (\alpha,\delta)g_2 &= (\gamma,\delta) \end{split}\]by simply using the transitivity of \(G_\alpha,G_\delta\), hence
\[(\alpha,\beta)g_1g_2 = (\gamma,\delta)\]and thus \(G\) is \(2\)-transitive, as desired.
\[\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\blacksquare\]Using this result, we can prove an auxiliary lemma that will be quite useful.
Lemma 1: If \(G \neq 1\) is a \(2\)-transitive group acting on \(X\), then the stabilizer \(G_\alpha\) of any element \(\alpha \in X\) is a maximal subgroup of \(G\).
Proof: Let \(g \in G \setminus G_\alpha\) be an element of the group that is not in the stabilizer, i.e., \(\alpha g = \beta \neq \alpha\), and let \(H = \langle G_\alpha,g \rangle\). We will show that \(H = G\), and since this will be true for any \(g \in G \setminus G_\alpha\), it will follow that \(G_\alpha\) is maximal. We know that the lateral classes of \(G_\alpha\) partition \(G\), hence let \(h \in G\) be an element of the group, and consider the lateral class \(G_\alpha h\). Let \(\beta h = \gamma\), and note that since \(G_\alpha\) is transitive on \(X \setminus \{\alpha\}\), we can find some \(g_\gamma \in G_\alpha\) s.t. \(\beta g_\gamma = \gamma\), hence \(\alpha gg_\gamma = \beta\). We claim that \(G_\alpha gg_\gamma = G_\alpha h\). Indeed, note that
\[\begin{split} (fgg_\gamma h^{-1})h &\in G_\alpha h\\ (fhg_\gamma^{-1}g^{-1})gg_\gamma &\in G_\alpha gg_\gamma \end{split}\]for any \(f \in G_\alpha\), and thus the claim follows. This shows that any lateral class of \(G_\alpha\) is contained in \(H\), implying that \(G \subseteq H\), and hence \(H = G\), as desired.
\[\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\blacksquare\]We can use the previous Lemma to show that \(S_n\) is a maximal subgroup of \(S_{n+1}\), and similarly that \(A_n\) is a maximal subgroup of \(A_{n+1}\). First, note that \(S_n\) acts transitively on \(\{1,...,n\}\), since we can take any element \(i\) to some other element \(j\) via the transposition \((i,j)\), and similarly \(A_n\) is also transitive since we can take \(i\) to \(j\) via \((i,k)(k,j)\). Now if we consider any fixed element \(i \in \{1,...,n+1\}\), then the stabilizer of this element in \(S_{n+1}\) is a copy of \(S_n\), hence it is transitive on \(\{1,...,n+1\} \setminus \{i\}\), implying that \(S_{n+1}\) is \(2\)-transitive, and thus \(S_n \leq S_{n+1}\) is maximal. We can conclude that \(A_n \leq A_{n+1}\) analogously.
We are now ready to prove that \(A_n\) is simple if \(n \geq 5\). The strategy will be as follows: we start with showing that \(A_5\) is simple, by analyzing its conjugacy classes and finding a contradiction with the existence of a normal proper non-trivial subgroup. We then proceed inductively on \(n\), and use the previous results about transitivity to complete the proof.
First, recall from the previous post that the conjugacy classes of \(S_n\) are completely determined by the integer partitions of \(n\), and in turn we can completely determine the conjugacy classes of \(A_n\) by analyzing the centralizers of these classes. Below is a table of the conjugacy classes of \(S_5\), where each row is indexed by the respective partition, \(x\) denotes a representative of the conjugacy class, \(C_{S_5}(x)\) denotes its centralizer, and \(x^{S_5}\) denotes its orbit.
| Partition | \(x\) | \(C_{S_5}(x)\) | \(\vert x^{S_5}\vert\) | \(\vert C_{S_5}(x) \vert\) |
|---|---|---|---|---|
| \((1,1,1,1,1)\) | \(1\) | \(S_5\) | \(1\) | \(120\) |
| \((2,1,1,1)\) | \((1,2)\) | \(\langle (1,2),(3,4),(3,4,5)\rangle\) | \(10\) | \(12\) |
| \((2,2,1)\) | \((1,2)(3,4)\) | \(\langle (1,2),(1,3,2,4)\rangle\) | \(15\) | \(8\) |
| \((3,1,1)\) | \((1,2,3)\) | \(\langle (1,2,3),(4,5)\rangle\) | \(20\) | \(6\) |
| \((3,2)\) | \((1,2,3)(4,5)\) | \(\langle (1,2,3),(4,5)\rangle\) | \(20\) | 6 |
| \((4,1)\) | \((1,2,3,4)\) | \(\langle (1,2,3,4)\rangle\) | \(30\) | 4 |
| \((5)\) | \((1,2,3,4,5)\) | \(\langle (1,2,3,4,5)\rangle\) | \(24\) | \(5\) |
Now recall that \(A_5\) contains only even permutations, hence the conjugacy classes corresponding to the partitions \((2,1,1,1),(3,2),(4,1)\) will have empty intersections with \(A_5\). The centralizer of the class corresponding to \((2,2,1)\) is not contained in \(A_5\), and thus the conjugacy class is also a class in \(A_5\), and the same is true for \((3,1,1)\). The centralizer of \((5)\) on the other hand is contained in \(A_5\), thus the class splits into two disjoint conjugacy classes in \(A_5\). With these observations, we obtain the following table of the conjugacy classes of \(A_5\):
| \(x\) | \(C_{A_5}(x)\) | \(\vert x^{A_5}\vert\) | \(\vert C_{A_5}(x) \vert\) |
|---|---|---|---|
| \(1\) | \(A_5\) | \(1\) | \(60\) |
| \((1,2)(3,4)\) | \(\langle (1,2)(3,4),(1,3)(2,4)\rangle\) | \(15\) | \(4\) |
| \((1,2,3)\) | \(\langle (1,2,3)\rangle\) | \(20\) | \(3\) |
| \((1,2,3,4,5)\) | \(\langle (1,2,3,4,5)\rangle\) | \(12\) | \(5\) |
| \((1,2,3,5,4)\) | \(\langle (1,2,3,5,4)\rangle\) | \(12\) | \(5\) |
Now assume that \(N \trianglelefteq A_5\) is a normal subgroup, thus it must be the disjoint union of conjugacy classes in \(A_5\), hence \(\vert N \vert \mid 60\), and on the other hand there must be integers \(x_1,x_2,x_3\) such that
\[\vert N \vert = 15x_1 + 20x_2 + 12x_3\]where \(x_1,x_2 \in \{0,1\},x_3 \in \{0,1,2\}\) correspond to the possible choices of conjugacy classes. These two facts together with a simple inspection show that either \(N = A_5\) or \(N = 1\), and thus \(A_5\) is indeed simple. This will serve as the base case of our induction in the following theorem.
Theorem 1: The alternating group \(A_n\) is simple for \(n \geq 5\).
Proof: We proceed via induction on \(n\). The base case of \(n = 5\) was already verified, so assume that the claim holds for values less than \(n+1\) and consider \(A_{n+1}\). Let \(i \in \{1,2,...,n+1\}\), and note that the stabilizer \(G_i\) is isomorphic to \(A_n\), and since \(A_{n+1}\) is transitive, it follows that all \(G_i\)’s are conjugate.
Let \(N \trianglelefteq A_{n+1}\) be a normal subgroup, and consider \(N \cap G_1 \trianglelefteq G_1\), and since \(G_1\) is isomorphic to \(A_n\), we know by the inductive hypothesis that either \(N \cap G_1 = G_1\) or \(N \cap G_1 = 1\). If \(N \cap G_1 = G_1\), since the \(G_i\)’s are conjugate and \(N\) is normal, it follows that \(G_i \subseteq N\) for all \(i\). However, note that \(N\) is normal but \(G_1\) isn’t, hence \(N\) properly contains each \(G_1\), and since each \(G_i\) is a maximal subgroup of \(A_{n+1}\), this implies that \(N = A_{n+1}\). Now assume that for all \(i\), we have \(N \cap G_i = 1\), and assume that \(N \neq 1\). Since \(G_1\) is maximal, it follows that \(NG_1 = A_{n+1}\), hence
\[\vert NG_1 \vert = \frac{\vert N \vert\vert G_1 \vert}{\vert N \cap G_1 \vert} = \frac{\vert N \vert}{\vert A_n\vert}\]hence \(\vert N \vert = n+1\). Let \(\sigma \in N \setminus\{1\}\), and write its decomposition as a product of disjoint cycles as follows:
\[\sigma = \sigma_1\ldots\sigma_k\]where each cycle has length \(c_i\), with \(c_1 \geq \ldots \geq c_k\). If \(c_1 \geq 3\), we may assume WLOG that \(\sigma_1 = (1,2,3,...,m)\), for some \(m \geq 3\), thus if we take \(\rho = (3,4,5)\), we get that \(\sigma^\rho\sigma^{-1}\) is an element of \(G_1 \cap N \setminus \{1\}\), which is a contradiction. If \(c_1 = 2\), then we may assume WLOG that \(\sigma_1 = (1,2)(3,4)\), and thus if we take \(\rho = (4,5,6)\) we again get that \(\sigma^\rho\sigma^{-1}\) is an element of \(G_1 \cap N \setminus \{1\}\), which is also a contradiction. This implies that \(A_{n+1}\) is simple, as desired.
\[\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\blacksquare\]The simplicity of \(A_n\) for \(n \geq 5\) also implies that \(A_n\) is the only proper non-trivial normal subgroup of \(S_n\) in these cases.
Theorem 2: If \(n \geq 5\), then \(A_n\) is the only proper non-trivial normal subgroup of \(S_n\).
Proof: Suppose that \(N \trianglelefteq S_n\) is a proper normal subgroup of \(S_n\) that is not equal to \(1\). Since \(A_n\) is also a normal subgroup, it follows that \(A_n \cap N\) is also normal and proper in \(S_n\). If \(A_n \cap N = 1\), then \(N\) must be composed only of odd permutations, however the product of two odd permutations is even, and thus \(A_n \cap N \neq 1\), i.e., \(A_n \cap N\) is a non-trivial normal subgroup of both \(A_n\) and \(N\), but since \(A_n\) is simple if \(n \geq 5\), it follows that \(A_n \cap N = A_n\), and thus \(A_n \subseteq N\). On the other hand, \(A_n\) is a maximal subgroup of \(S_n\), thus it follows that \(N = A_n\), as desired.
\[\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\blacksquare\]Now it is actually fairly straightforward to check that \(S_n\) is not solvable for \(n \geq 5\). Recall that a group \(G\) is solvable iff there exists some normal series
\[G = G_0 \trianglerighteq G_1 \ldots \trianglerighteq G_k = 1\]where each \(G_i \trianglelefteq G\), and \(G_i/G_{i+1}\) is abelian. If \(n \geq 5\), since neither \(S_n\) nor \(A_n\) are abelian, and since the only normal subgroup of \(S_n\) is \(A_n\), any normal series is of the form
\[S_n \trianglerighteq A_n \trianglerighteq 1\]hence the series does not satisfy the solvability conditions, implying that \(S_n\) is not solvable if \(n \geq 5\).
On the other hand, the groups \(S_1,S_2,S_3,S_4\) are all solvable. This is immediately true for \(S_1\) and \(S_2\), as they are abelian. The solvability of \(S_3\) follows from a more general fact: the dihedral group \(D_n\) is actually solvable for all \(n\), since if we consider the subgroup of all rotations \(C_n\), it has order \(n\), and since \(D_n\) has order \(2n\) it follows that the index of \(C_n\) is two, and thus it is normal, hence the series
\[D_n \trianglerighteq C_n \trianglerighteq 1\]is normal, and \(D_n/C_n\) is abelian as it is a group of order two, and since \(S_3 = D_3\), the result follows. Now in order to see that \(S_4\) is solvable, it will require some more effort. We first consider the set
\[\begin{split} X &= \{X_1,X_2,X_3\},\\ X_1 &= \{\{1,2\},\{3,4\}\},\\ X_2 &= \{\{1,3\},\{2,4\}\},\\ X_3 &= \{\{1,4\},\{2,3\}\} \end{split}\]and note that \(S_4\) acts on \(X\) as follows by permuting the elements of the sets, e.g.,
\[\begin{split} \{1,2\}(1,2,3) &= \{1(1,2,3),2(1,2,3)\} = \{2,3\}\\ X_1(1,2,3) &= X_3 \end{split}\]Since an action is nothing more than an homomorphism from the group to the group of symmetries of the set, it follows that we obtain an homomorphism \(\varphi:S_4 \mapsto S_3\), since \(\text{Sym}(X) = S_3\). It follows by a simple computation that
\[\text{ker}(\varphi) = \{1,(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\}\]and the set \(\text{ker}(\varphi)\) is often called the Klein group \(K_4\) on \(4\) elements. From this, we have that \(S_4/K_4\) is isomorphic to some subgroup of \(S_3\), but on the other hand \(S_4/K_4\) has order \(6\), thus
\[S_4/K_4 \cong S_3\]Now note that \(K_4\) has \(4\) elements, and thus it is abelian, and in particular it is solvable, and since it is the kernel of an homomorphism it follows that \(K_4\) is a normal solvable subgroup of \(S_4\). Recall that if \(N\) is a normal solvable subgroup of \(G\), and if \(G/N\) is solvable, then \(G\) is also solvable, thus this implies that \(S_4\) is solvable.