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The 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, where p is prime.

The 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, \({\mathbb Z}_p\), 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, \(A_n\), is simple for \(n \geq 5\). The proof of this result requires several lemmas.

Lemma

Let \(N\) be a normal subgroup of \(A_n\), where \(n \geq 3\). If \(N\) contains a \(3\)-cycle, then \(N = A_n\).

Proof

We will first show that \(A_n\) is generated by \(3\)-cycles of the specific form \((i,j,k)\), 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 \[\begin{aligned}(i, a, j) & = (i, j, a)^2 \\ (i, a, b) & = (i, j, b) (i, j, a)^2 \\ (j, a, b) & = (i, j, b)^2 (i, j, a) \\ (a, b, c) & = (i, j, a)^2 (i, j, c) (i, j, b)^2 (i, j, a)\end{aligned}\]. 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 \((i, j, a)\). Using the normality of \(N\), we see that \[\begin{aligned}\end{aligned}\] is in \(N\). Hence, \(N\) must contain all of the \(3\)-cycles \((i, j, k)\) for \(1 \leq k \leq n\). By , these \(3\)-cycles generate \(A_n\); hence, \(N = A_n\).

Lemma

For \(n \geq 5\), every nontrivial normal subgroup \(N\) of \(A_n\) contains a \(3\)-cycle.

Proof

Let \(\sigma\) be an arbitrary element in a normal subgroup \(N\). There are several possible cycle structures for \(\sigma\).

\(\sigma\) is a \(3\)-cycle.
\(\sigma\) is the product of disjoint cycles, \(\sigma = \tau(a_1, a_2, \ldots, a_r) \in N\), where \(r \gt 3\).
\(\sigma\) is the product of disjoint cycles, \(\sigma = \tau(a_1, a_2, a_3)(a_4, a_5, a_6)\).
\(\sigma = \tau(a_1, a_2, a_3)\), where \(\tau\) is the product of disjoint 2-cycles.
\(\sigma = \tau (a_1, a_2) (a_3, a_4)\), 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, \(\sigma\), and at least one of these cycles has length greater than 3, say \(\sigma = \tau(a_1, a_2, \ldots, a_r)\), then \[\begin{aligned}\end{aligned}\] is in \(N\) since \(N\) is normal; hence, \[\begin{aligned}\end{aligned}\] is also in \(N\). Since \[\begin{aligned}\sigma^{-1}(a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1} & = \sigma^{-1}(a_1, a_2, a_3)\sigma(a_1, a_3, a_2) \\ & = (a_1, a_2, \ldots, a_r)^{-1}\tau^{-1}(a_1, a_2, a_3) \tau(a_1, a_2, \ldots, a_r)(a_1, a_3, a_2) \\ & = (a_1, a_r, a_{r-1}, \ldots, a_2 )(a_1, a_2, a_3) (a_1, a_2, \ldots, a_r)(a_1, a_3, a_2) \\ & = (a_1, a_3, a_r)\end{aligned}\], \(N\) must contain a \(3\)-cycle; hence, \(N = A_n\).

Now suppose that \(N\) contains a disjoint product of the form \[\begin{aligned}\end{aligned}\]. Then \[\begin{aligned}\end{aligned}\] since \[\begin{aligned}\end{aligned}\]. So \[\begin{aligned}\sigma^{-1}(a_1, a_2, a_4) \sigma(a_1, a_2, a_4)^{-1} & = [ \tau (a_1, a_2, a_3) (a_4, a_5, a_6) ]^{-1} (a_1, a_2, a_4) \tau (a_1, a_2, a_3) (a_4, a_5, a_6) (a_1, a_2, a_4)^{-1} \\ & = (a_4, a_6, a_5) (a_1, a_3, a_2) \tau^{-1}(a_1, a_2, a_4) \tau (a_1, a_2, a_3) (a_4, a_5, a_6) (a_1, a_4, a_2) \\ & = (a_4, a_6, a_5)(a_1, a_3, a_2) (a_1, a_2, a_4) (a_1, a_2, a_3) (a_4, a_5, a_6)(a_1, a_4, a_2) \\ & = (a_1, a_4, a_2, a_6, a_3)\end{aligned}\]. 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 \(\sigma = \tau(a_1, a_2, a_3)\), where \(\tau\) is the product of disjoint \(2\)-cycles. Since \(\sigma \in N\), \(\sigma^2 \in N\), and \[\begin{aligned}\sigma^2 & = \tau(a_1, a_2, a_3)\tau(a_1, a_2, a_3) \\ & =(a_1, a_3, a_2)\end{aligned}\]. So \(N\) contains a \(3\)-cycle.

The only remaining possible case is a disjoint product of the form \[\begin{aligned}\end{aligned}\], where \(\tau\) is the product of an even number of disjoint \(2\)-cycles. But \[\begin{aligned}\end{aligned}\] is in \(N\) since \((a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1}\) is in \(N\); and so \[\begin{aligned}\sigma^{-1}(a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1} & = \tau^{-1} (a_1, a_2) (a_3, a_4) (a_1, a_2, a_3) \tau (a_1, a_2)(a_3, a_4)(a_1, a_2, a_3)^{-1} \\ & = (a_1, a_3)(a_2, a_4)\end{aligned}\]. Since \(n \geq 5\), we can find \(b \in \{1, 2, \ldots, n \}\) such that \(b \neq a_1, a_2, a_3, a_4\). Let \(\mu = (a_1, a_3, b)\). Then \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\mu^{-1} (a_1, a_3)(a_2, a_4) \mu (a_1, a_3)(a_2, a_4) & = (a_1, b a_3)(a_1, a_3)(a_2, a_4) (a_1, a_3, b)(a_1, a_3)(a_2, a_4) \\ & = (a_1 a_3 b )\end{aligned}\]. Therefore, \(N\) contains a \(3\)-cycle. This completes the proof of the lemma.

Sage can easily determine if a subgroup is normal or not. If so, it can create the quotient group. However, the construction creates a new permuation group, isomorphic to the quotient group, so its utility is limited.

Historical 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.

Symbols used here

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

\sigma,\ s,\ \sigma^2

standard deviation, sample s.d., variance
Typical distance from the mean; its square.

\bar{x},\ \mu

sample mean, population mean
Average of the data; average of the whole population.

imaginary unit
i² = −1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

\blacksquare\ \text{or}\ \square

end of proof (halmos)
Marks the point where the statement has been established.

a \equiv b \pmod n

congruent modulo n
n divides a − b; a and b have the same remainder.

a \mid b,\ \gcd(a,b)

divides, greatest common divisor
b is a multiple of a; the largest number dividing both.

(G, \cdot),\ e,\ g^{-1}

group, identity, inverse
A set with an operation; the do-nothing element; the element that undoes g.

G \cong H,\ G / N

isomorphic, quotient group
Same structure; the group of cosets of a normal subgroup N.

\mathbb{Z}/n\mathbb{Z},\ \mathbb{Z}_n

integers modulo n
The remainders 0…n−1 with clock arithmetic.

\operatorname{Hom}(A, B),\ f \circ g

arrows from A to B, composition
The set of morphisms; do g then f.

Questions people ask

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group — that is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

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Parts of this page are adapted from Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Condensed and re-explained here; errors are ours.

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