Section10.2The Simplicity of the Alternating Group

Of special interest are groups with no nontrivial normal subgroups. Such groups are called simple groups. Of course, we already have a whole class of examples of simple groups, $${\mathbb Z}_p\text{,}$$ where $$p$$ is prime. These groups are trivially simple since they have no proper subgroups other than the subgroup consisting solely of the identity. Other examples of simple groups are not so easily found. We can, however, show that the alternating group, $$A_n\text{,}$$ is simple for $$n \geq 5\text{.}$$ The proof of this result requires several lemmas.

To show that the $$3$$-cycles generate $$A_n\text{,}$$ we need only show that any pair of transpositions can be written as the product of $$3$$-cycles. Since $$(a, b) = (b, a)\text{,}$$ every pair of transpositions must be one of the following:

\begin{align*} (a,b)(a,b) & = \identity\\ (a,b)(c,d) & = (a,c,b)(a,c,d)\\ (a,b)(a,c) & = (a,c,b)\text{.} \end{align*}

We will first show that $$A_n$$ is generated by $$3$$-cycles of the specific form $$(i,j,k)\text{,}$$ where $$i$$ and $$j$$ are fixed in $$\{ 1, 2, \ldots, n \}$$ and we let $$k$$ vary. Every $$3$$-cycle is the product of $$3$$-cycles of this form, since

\begin{align*} (i, a, j) & = (i, j, a)^2\\ (i, a, b) & = (i, j, b) (i, j, a)^2\\ (j, a, b) & = (i, j, b)^2 (i, j, a)\\ (a, b, c) & = (i, j, a)^2 (i, j, c) (i, j, b)^2 (i, j, a)\text{.} \end{align*}

Now suppose that $$N$$ is a nontrivial normal subgroup of $$A_n$$ for $$n \geq 3$$ such that $$N$$ contains a $$3$$-cycle of the form $$(i, j, a)\text{.}$$ Using the normality of $$N\text{,}$$ we see that

\begin{equation*} [(i, j)(a, k)](i, j, a)^2 [(i, j)(a, k)]^{-1} = (i, j, k) \end{equation*}

is in $$N\text{.}$$ Hence, $$N$$ must contain all of the $$3$$-cycles $$(i, j, k)$$ for $$1 \leq k \leq n\text{.}$$ By Lemma 10.8, these $$3$$-cycles generate $$A_n\text{;}$$ hence, $$N = A_n\text{.}$$

Let $$\sigma$$ be an arbitrary element in a normal subgroup $$N\text{.}$$ There are several possible cycle structures for $$\sigma\text{.}$$

• $$\sigma$$ is a $$3$$-cycle.

• $$\sigma$$ is the product of disjoint cycles, $$\sigma = \tau(a_1, a_2, \ldots, a_r) \in N\text{,}$$ where $$r \gt 3\text{.}$$

• $$\sigma$$ is the product of disjoint cycles, $$\sigma = \tau(a_1, a_2, a_3)(a_4, a_5, a_6)\text{.}$$

• $$\sigma = \tau(a_1, a_2, a_3)\text{,}$$ where $$\tau$$ is the product of disjoint 2-cycles.

• $$\sigma = \tau (a_1, a_2) (a_3, a_4)\text{,}$$ where $$\tau$$ is the product of an even number of disjoint 2-cycles.

If $$\sigma$$ is a $$3$$-cycle, then we are done. If $$N$$ contains a product of disjoint cycles, $$\sigma\text{,}$$ and at least one of these cycles has length greater than 3, say $$\sigma = \tau(a_1, a_2, \ldots, a_r)\text{,}$$ then

\begin{equation*} (a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1} \end{equation*}

is in $$N$$ since $$N$$ is normal; hence,

\begin{equation*} \sigma^{-1}(a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1} \end{equation*}

is also in $$N\text{.}$$ Since

\begin{align*} \sigma^{-1}(a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1} & = \sigma^{-1}(a_1, a_2, a_3)\sigma(a_1, a_3, a_2)\\ & = (a_1, a_2, \ldots, a_r)^{-1}\tau^{-1}(a_1, a_2, a_3) \tau(a_1, a_2, \ldots, a_r)(a_1, a_3, a_2)\\ & = (a_1, a_r, a_{r-1}, \ldots, a_2 )(a_1, a_2, a_3) (a_1, a_2, \ldots, a_r)(a_1, a_3, a_2)\\ & = (a_1, a_3, a_r)\text{,} \end{align*}

$$N$$ must contain a $$3$$-cycle; hence, $$N = A_n\text{.}$$

Now suppose that $$N$$ contains a disjoint product of the form

\begin{equation*} \sigma = \tau(a_1, a_2, a_3)(a_4, a_5, a_6)\text{.} \end{equation*}

Then

\begin{equation*} \sigma^{-1}(a_1, a_2, a_4)\sigma(a_1, a_2, a_4)^{-1} \in N \end{equation*}

since

\begin{equation*} (a_1, a_2, a_4)\sigma(a_1, a_2, a_4)^{-1} \in N\text{.} \end{equation*}

So

\begin{align*} \sigma^{-1}(a_1, a_2, a_4) \sigma(a_1, a_2, a_4)^{-1} & = [ \tau (a_1, a_2, a_3) (a_4, a_5, a_6) ]^{-1} (a_1, a_2, a_4) \tau (a_1, a_2, a_3) (a_4, a_5, a_6) (a_1, a_2, a_4)^{-1}\\ & = (a_4, a_6, a_5) (a_1, a_3, a_2) \tau^{-1}(a_1, a_2, a_4) \tau (a_1, a_2, a_3) (a_4, a_5, a_6) (a_1, a_4, a_2)\\ & = (a_4, a_6, a_5)(a_1, a_3, a_2) (a_1, a_2, a_4) (a_1, a_2, a_3) (a_4, a_5, a_6)(a_1, a_4, a_2)\\ & = (a_1, a_4, a_2, a_6, a_3)\text{.} \end{align*}

So $$N$$ contains a disjoint cycle of length greater than 3, and we can apply the previous case.

Suppose $$N$$ contains a disjoint product of the form $$\sigma = \tau(a_1, a_2, a_3)\text{,}$$ where $$\tau$$ is the product of disjoint 2-cycles. Since $$\sigma \in N\text{,}$$ $$\sigma^2 \in N\text{,}$$ and

\begin{align*} \sigma^2 & = \tau(a_1, a_2, a_3)\tau(a_1, a_2, a_3)\\ & =(a_1, a_3, a_2)\text{.} \end{align*}

So $$N$$ contains a $$3$$-cycle.

The only remaining possible case is a disjoint product of the form

\begin{equation*} \sigma = \tau (a_1, a_2) (a_3, a_4)\text{,} \end{equation*}

where $$\tau$$ is the product of an even number of disjoint $$2$$-cycles. But

\begin{equation*} \sigma^{-1}(a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1} \end{equation*}

is in $$N$$ since $$(a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1}$$ is in $$N\text{;}$$ and so

\begin{align*} \sigma^{-1}(a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1} & = \tau^{-1} (a_1, a_2) (a_3, a_4) (a_1, a_2, a_3) \tau (a_1, a_2)(a_3, a_4)(a_1, a_2, a_3)^{-1}\\ & = (a_1, a_3)(a_2, a_4)\text{.} \end{align*}

Since $$n \geq 5\text{,}$$ we can find $$b \in \{1, 2, \ldots, n \}$$ such that $$b \neq a_1, a_2, a_3, a_4\text{.}$$ Let $$\mu = (a_1, a_3, b)\text{.}$$ Then

\begin{equation*} \mu^{-1} (a_1, a_3)(a_2, a_4) \mu (a_1, a_3)(a_2, a_4) \in N \end{equation*}

and

\begin{align*} \mu^{-1} (a_1, a_3)(a_2, a_4) \mu (a_1, a_3)(a_2, a_4) & = (a_1, b a_3)(a_1, a_3)(a_2, a_4) (a_1, a_3, b)(a_1, a_3)(a_2, a_4)\\ & = (a_1 a_3 b )\text{.} \end{align*}

Therefore, $$N$$ contains a $$3$$-cycle. This completes the proof of the lemma.

Let $$N$$ be a normal subgroup of $$A_n\text{.}$$ By Lemma 10.10, $$N$$ contains a $$3$$-cycle. By Lemma 10.9, $$N = A_n\text{;}$$ therefore, $$A_n$$ contains no proper nontrivial normal subgroups for $$n \geq 5\text{.}$$

SubsectionHistorical Note

One of the foremost problems of group theory has been to classify all simple finite groups. This problem is over a century old and has been solved only in the last few decades of the twentieth century. In a sense, finite simple groups are the building blocks of all finite groups. The first nonabelian simple groups to be discovered were the alternating groups. Galois was the first to prove that $$A_5$$ was simple. Later, mathematicians such as C. Jordan and L. E. Dickson found several infinite families of matrix groups that were simple. Other families of simple groups were discovered in the 1950s. At the turn of the century, William Burnside conjectured that all nonabelian simple groups must have even order. In 1963, W. Feit and J. Thompson proved Burnside's conjecture and published their results in the paper “Solvability of Groups of Odd Order,” which appeared in the Pacific Journal of Mathematics. Their proof, running over 250 pages, gave impetus to a program in the 1960s and 1970s to classify all finite simple groups. Daniel Gorenstein was the organizer of this remarkable effort. One of the last simple groups was the “Monster,” discovered by R. Greiss. The Monster, a $$196{,}833 \times 196{,}833$$ matrix group, is one of the 26 sporadic, or special, simple groups. These sporadic simple groups are groups that fit into no infinite family of simple groups. Some of the sporadic groups play an important role in physics.