CHAPTER 13 · MA-0561

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Simple Groups

Simple Groups

Definition (Simple group)

Let \(G\) be a non-trivial group. We say that \(G\) is simple if it has no proper normal subgroups, i.e., if \(H \triangleleft G\), then \(H = \{ 1_{G} \}\) or \(H=G\).

Theorem (Simple abelian groups are \(\mathbb{Z}_{p}\))

Let \(G\) be an abelian group. Then \(G\) is simple if and only if \(G=\mathbb{Z}_{p}\) with \(p\) prime.

Proof: (\(\impliedby\)): This direction is straightforward, since it suffices to apply Lagrange to note that \(G\) has no proper subgroups. (\(\implies\)): Since \(G\) is abelian, every subgroup is normal. Moreover, since \(G\) is simple, \(G \neq \{ 1_{G} \}\). Then there exists \(x \in G\) such that \((x \neq 1_{G})\). We have that

\[\langle x \rangle \triangleleft G \implies \langle x \rangle = G \implies G \text{ is cyclic} \implies G \cong \mathbb{Z} \quad \lor \quad G \cong \mathbb{Z}_{n}, \text{ for some } n \in \mathbb{N}.\]

Since \(G\) is simple, \(G \not\cong \mathbb{Z}\), so \(G \cong \mathbb{Z}_{n}\) for some \(n \in \mathbb{N}\). Moreover, since \(G\) is simple, \(n\) is prime, from which we conclude the result.

Reminder (Sign and alternating group)

If \(\sigma \in S_{n}\) is such that \(\sigma = \beta_{1} \cdots \beta_{t}\) is its decomposition into disjoint cycles of \(\sigma\), then \(\operatorname{sgn}( \sigma ) := (-1)^{n-t}\). If \(\operatorname{sgn}( \sigma ) = 1\) we say that \(\sigma\) is even and \(A_{n} = \{ \sigma \in S_{n}, \sigma \text{ is even} \}.\)

Exercises (On \(3\)-cycles)

  1. \(A_{n}\) is generated by the 3-cycles if \(n\geq {5}\).
  2. In \(S_{5}\), all 3-cycles are conjugate.
  3. In \(A_{5}\), all 3-cycles are conjugate.

Theorem (\(A_{5}\) is simple)

\(A_{5}\) is simple

Proof: Let \(H \neq \{ \mathrm{id}_{S_{5}} \}\) be such that \(H \triangleleft A_{5}\). By the previous exercises, it suffices to show that \(H\) contains a 3-cycle, since if \(\sigma \in H\) is a 3-cycle, then \(\sigma^{A_{5}} \subseteq H\) because \(H \triangleleft A_{5}\), hence \(\langle \sigma^{A_{5}} \rangle = \{ \tau: \tau \text{ is a 3-cycle} \} \subseteq H\) and therefore \(H = A_{5}\).

How many possible configurations are there for the decomposition into disjoint cycles?

  1. \(\sigma\) decomposes into 5 1-cycles, and therefore \(\sigma \in A_{5}\).
  2. \(\sigma\) decomposes into one 2-cycle and 3 1-cycles, and therefore \(\sigma \not\in A_{5}\).
  3. \(\sigma\) decomposes into one 3-cycle and 2 1-cycles, and therefore \(\sigma \in A_{5}\).
  4. \(\sigma\) decomposes into one 4-cycle and 1 1-cycle, and therefore \(\sigma \not\in A_{5}\).
  5. \(\sigma\) decomposes into one 5-cycle, and therefore \(\sigma \in A_{5}\).
  6. \(\sigma\) decomposes into two 2-cycles and one 1-cycle, and therefore \(\sigma \in A_{5}\).
  7. \(\sigma\) decomposes into one 3-cycle and one 2-cycle, and therefore \(\sigma \not\in A_{5}\).

Hence, if \(\sigma \in H \setminus \{ 1_{G} \}\), its factorisation is of type 3, 5 or 6. If \(\sigma\) is of type 3, then it contains a 3-cycle and therefore the result holds. If \(\sigma\) is of type 5, suppose without loss of generality that \(\sigma = \begin{pmatrix}1 & 2 & 3 & 4 & 5\end{pmatrix}\). Let \(\alpha = \begin{pmatrix}1 & 2 & 3\end{pmatrix} \in A_{5}\). Hence,

\[w = \begin{pmatrix} 2 & 3 & 1 & 4 & 5 \end{pmatrix} =\alpha \sigma \alpha^{-1} \in H, \text{ since }H \triangleleft A_{5} \implies w \sigma ^{-1} = \begin{pmatrix} 1 & 2 & 4 \end{pmatrix} \in H.\]

If \(\sigma\) is of type 6, suppose without loss of generality that \(\sigma = \begin{pmatrix}1 & 2\end{pmatrix}\begin{pmatrix}3 & 4\end{pmatrix}\). Let \(\beta = \begin{pmatrix}3 & 4 & 5\end{pmatrix} \in A_{5}\). Note that \(\delta :=\beta \sigma \beta ^{-1} = \begin{pmatrix}1 & 2\end{pmatrix}\begin{pmatrix}4 & 5\end{pmatrix} \in H\). Finally, \(\sigma \cdot \delta = \begin{pmatrix}3 & 4 & 5\end{pmatrix} \in H\). Conclude that \(A_{5}\) is simple.

Theorem (\(A_{6}\) is simple)

\(A_{6}\) is simple

Proof: Let \(H\neq \{ \mathrm{id}_{S_{6}} \}\) be such that \(H \triangleleft A_{6}\). Suppose for contradiction that if \(\sigma \in H \setminus \{ \mathrm{id}_{S_{6}} \}\), then \(\sigma\) moves every element of \(\{ 1,2,\dots,6 \}\). How many possible configurations are there for the decomposition into disjoint cycles?

  1. \(\sigma\) decomposes into three 2-cycles, and therefore \(\sigma \not\in A_{6}\).
  2. \(\sigma\) decomposes into one 2-cycle and one 4-cycle, and therefore \(\sigma \in A_{6}\).
  3. \(\sigma\) decomposes into two 3-cycles, and therefore \(\sigma \in A_{6}\).

If \(\sigma\) is of type 2, consider \(\sigma = \begin{pmatrix}1 & 2\end{pmatrix}\begin{pmatrix}3 & 4 & 5 & 6\end{pmatrix}\). Since \(H\) is a subgroup, if \(\sigma \in H\), then \(\sigma^{2} \in H\). But \(\sigma^{2}(1) = 1\), a contradiction. On the other hand, if \(\sigma\) is of type 3, consider \(\sigma=\begin{pmatrix}1 & 2 & 3\end{pmatrix}\begin{pmatrix}4 & 5 & 6\end{pmatrix}\). Hence we have that \(\sigma ^{-1} = \begin{pmatrix}3 & 2 & 1\end{pmatrix}\begin{pmatrix}6 & 5 & 4\end{pmatrix}\). Let \(\gamma = \begin{pmatrix}2 & 3 & 4\end{pmatrix}\). Since \(H\) is a subgroup, we have that \(\sigma\gamma \sigma ^{-1} \gamma ^{-1} = \begin{pmatrix}1 & 5 & 3 & 2 & 4\end{pmatrix} \in H\), a contradiction. Conclude that there exist \(\sigma \in H\) and \(i \in \{ 1,\dots,6 \}\) such that \(\sigma(i) = i\). Let \(F = \{ \sigma \in A_{6}: \sigma(i)=i \}\). Note that \(F \neq \{ 1_{S_{6}} \}\) and \(F \cap H \neq \{ 1_{S_{6}} \}\). It is easy to see that \(F \cong A_{5}\). Finally, since \(F \cap H \triangleleft F \cong A_{5}\) and \(A_{5}\) is simple, we have that \(F \cap H = F\), hence \(F \subseteq H\). Conclude that \(A_{6}\) is simple because it contains a 3-cycle.

Theorem (\(A_{n}\) is simple for \(n \geq 5\))

\(A_{n}\) is simple for \(n \geq5\)

Proof: Consider the case \(n>6\). Let \(H\neq \{ \mathrm{id}_{S_{n}} \}\) be such that \(H \triangleleft A_{n}\) Let \(\sigma \in H \setminus \{ \mathrm{id}_{Sn} \}\). Then there exists \(i \in \{ 1,\dots,n \}\) such that \(\sigma(i) = j \neq i\). Let \(\alpha\) be a 3-cycle such that \(\alpha(i)=i\) and \(\alpha(j)\neq j\). Note that

\[\alpha(\sigma(i)) = \alpha(j) \neq j \quad \land \quad\sigma(\alpha(i)) = \sigma(i) =j \implies \alpha \sigma \neq \sigma\alpha \implies \alpha\sigma\alpha ^{-1}\sigma ^{-1} \in H \setminus \{1_{S+n}\}.\]

We know that conjugation of permutations does not change the cycle structure. Hence \(\sigma \alpha ^{-1} \sigma ^{-1}\) is a 3-cycle and \(\gamma = \alpha\sigma\alpha ^{-1}\sigma ^{-1} \in H\) is a product of two 3-cycles and therefore moves at most 6 elements. Let \(i_{1},\dots i_{k}\), with \(k \leq 6\), be the elements that \(\gamma\) moves. Now, let \(F:=\{ \sigma \in A_{n}: \sigma(\ell) = \ell \quad \forall \ell \not\in \{ i_{1},\dots,i_{k} \}\}\). It is easy to see that \(F \cong S_{6}\). and \(F \cap H \neq \{ 1_{S_{n}} \}\). Thus, \(F \cap H \triangleleft F \cong A_{6}\), so \(F \cap H = F\) and therefore \(F \subseteq H\). In particular, \(\begin{pmatrix}i_{1} & i_{2} & i_{3}\end{pmatrix} \in F \subseteq H\), from which we conclude that it is simple because it contains a \(3-\)cycle..