Lemma10.8
The alternating group $A_n$ is generated by $3$-cycles for $n \geq 3\text{.}$
📚 The CoCalc Library - books, templates and other resources
License: OTHER
Kernel:
In [ ]:
Important: to view this notebook properly you will need to execute the cell above, which assumes you have an Internet connection. It should already be selected, or place your cursor anywhere above to select. Then press the "Run" button in the menu bar above (the right-pointing arrowhead), or press Shift-Enter on your keyboard.
ParseError: KaTeX parse error: \newcommand{\lt} attempting to redefine \lt; use \renewcommand
Section10.2The Simplicity of the Alternating Group
¶Of special interest are groups with no nontrivial normal subgroups. Such groups are called . 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.
Proof
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*}
(ab)(ab) & = \identity\\
(ab)(cd) & = (acb)(acd)\\
(ab)(ac) & = (acb).
\end{align*}
Lemma10.9
Let $N$ be a normal subgroup of $A_n\text{,}$ where $n \geq 3\text{.}$ If $N$ contains a $3$-cycle, then $N = A_n\text{.}$
Proof
We will first show that $A_n$ is generated by 3-cycles of the specific form $(ijk)\text{,}$ 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).
\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, $N = A_n\text{.}$
Lemma10.10
For $n \geq 5\text{,}$ every nontrivial normal subgroup $N$ of $A_n$ contains a $3$-cycle.
Proof
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, $\sigma = \tau(a_1 a_2 \cdots a_r) \in N\text{,}$ 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 $\sigma = \tau(a_1 a_2 \cdots a_r)\text{,}$ 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 \cdots a_r)^{-1}\tau^{-1}(a_1 a_2 a_3) \tau(a_1 a_2 \cdots a_r)(a_1 a_3 a_2)\\
& = (a_1 a_r a_{r-1} \cdots a_2 )(a_1 a_2 a_3) (a_1 a_2 \cdots a_r)(a_1 a_3 a_2)\\
& = (a_1 a_3 a_r),
\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).
\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.
\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).
\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).
\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),
\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).
\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 ).
\end{align*}
Therefore, $N$ contains a 3-cycle. This completes the proof of the lemma.
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.