3.2: Alternating Groups

Definition: Alternating groups

Let \(n \geq 2 \) be an integer. Then the alternating group \(A_n\) is the set of all even permutations in \(S_n\) under composition. 

\(A_n = \{ \sigma \in S_n \mid \sigma \text{ is an even permutation} \} \). 

\(|A_n|=\dfrac{n!}{2}\) for  \(n \geq 2 \).

For \(n=1\), \(A_1\) is trivial and has order one.

Example \(\PageIndex{1}\)

\( A_3=\{1, (1, 2, 3), (3, 2, 1)\}.  A_3\) is a cyclic group of order \(3.\)

Theorem \(\PageIndex{1}\)

For every integer \(n\geq 1\), the alternating group \(A_n\)  is a subgroup of the symmetric group \(S_n.\) That is, \(A_n\leq S_n.\)

Proof

Since \(A_n\subseteq S_n), we shall prove this using the subgroup test. 

The identity permutation \(e\) can be expressed
as a product of zero transpositions. Since zero is even, \(e\) is an even permutation. Therefore, \(e\in A_n.\)

Hence, \(A_n\neq\varnothing\).

Let \(\sigma,\tau\in A_n\).

Since \(\sigma\) and \(\tau\) are even permutations, there exist nonnegative integers \(r,s\) such that \(\sigma=\alpha_1\alpha_2\cdots\alpha_{2r}
\) and \(\tau=\beta_1\beta_2\cdots\beta_{2s},\) where each \(\alpha_i\) and \(\beta_j\) is a transposition.

Therefore, \(\sigma\tau=\alpha_1\cdots\alpha_{2r}\beta_1\cdots\beta_{2s}.\) This is a product of \(2r+2s=2(r+s) \) transpositions.

Since \(2(r+s)\) is even, \(\sigma\tau\) is an even permutation.

Thus,\(\sigma\tau\in A_n.\)

Hence, \(A_n\) is closed under composition.

Let \(\sigma\in A_n\). Since \(\sigma\) is even, we may write \( \sigma=\alpha_1\alpha_2\cdots\alpha_{2r},\) where each \(\alpha_i\) is a transposition.

Since every transposition is its own inverse \(\alpha_i^{-1}=\alpha_i,\)
\begin{align*}
\sigma^{-1}
&=(\alpha_1\alpha_2\cdots
\alpha_{2r})^{-1}\\
&=\alpha_{2r}^{-1}\cdots
\alpha_2^{-1}\alpha_1^{-1}\\
&=\alpha_{2r}\cdots
\alpha_2\alpha_1.
\end{align*}

Thus, \(\sigma^{-1}\) is a product of \(2r\) transpositions. Since \(2r\) is even, \(\sigma^{-1}\) is an even permutation.

Therefore, \( \sigma^{-1}\in A_n.\)

Hence, \(A_n\leq S_n.\)

Example \(\PageIndex{2}\)

Determine which of the following permutations belong to \(A_6\):
\(
(1234),\qquad
(123)(45),\qquad
(1234)(56),\qquad
(12)(34).
\)

Theorem \(\PageIndex{2}\)

\(A_n\)  is non-abelian for  integers \(n \ge 4\).

Proof:

We exhibit two elements of \(A_4\) that do not commute. Hence \(A_4\) is nonabelian. Since \(A_4\) embeds in \(A_n\) for every \(n\geq 4\), the group \(A_n\) is also nonabelian.

Note \(A_4=\{e, (12)(34), (13)(24), (14)(23), (123), (124), (134), (234), (432), (431), (421), (321) \}\). 

Let \(a=(1,2)(3,4)\) and \(b=(4,1)(4,2)\).

Consider \(ab=(1,2)(3,4)(4,1)(4,2)=(1,3,4)\).

Further consider, \(ba=(4,1)(4,2)(1,2)(3,4)=(4,3,2)\).

Since \((1,3,4) \ne (4,3,2),  A_4\) is non-abelian.

Since  \(A_4 \) embeds in \(A_n, \; \forall n \ge 4\), \(A_n\) is non-abelian for \(n\ge 4\). \(\square\)

Example \(\PageIndex{3}\)

Find a cyclic subgroup of \(A_8\) that has order 4.  

Answer

Consider \(
\sigma=(1\,2\,3\,4)(5\,6).
\) Since a \(4\)-cycle is odd and a transposition is odd, their product is even.
Thus, \(
\sigma\in A_8
\). Also, the order of 
\(
|\sigma|
=lcm(4,2)
=4.
\)

Therefore, the cyclic subgroup generated by \(\sigma\) is \(
\left\langle (1\,2\,3\,4)(5\,6) \right\rangle
\) and its elements are
\(
\left\{
e,\,
(1\,2\,3\,4)(5\,6),\,
(1\,3)(2\,4),\,
(1\,4\,3\,2)(5\,6)
\right\}.
\)

Hence,
\(
\left|\left\langle (1\,2\,3\,4)(5\,6) \right\rangle\right|
=4.
\)

Example \(\PageIndex{4}\)

 Consider \(\left\{e,(1\,2)(3\,4),\, (1\,3)(2\,4),\, (1\,4)(2\,3) \right\}\) which is a subgroup of \(A_8\) of order 4.

Let \( H=\left\{ e,\, (1\,2)(3\,4),\, (1\,3)(2\,4),\, (1\,4)(2\,3) \right\}. \) All three non-identity elements are products of two transpositions, so they are even. Hence, \( H \leq A_8. \) Also, every non-identity element has order \(2\). Therefore, \(H\) is not cyclic. 

Example \(\PageIndex{5}\)

Find the conjugacy classes of \(A_3\).

Solution

\(\{e\}, \{(1, 2, 3)\}, \{(3, 2, 1)\}.\)

\([e]=\{\sigma^{-1} e\sigma \mid \sigma \in A_3\}=\{e\}.\)

\([(1, 2, 3)]=\{\sigma^{-1} (1, 2, 3) \sigma \mid \sigma \in A_3\}= \{e^{-1} (1, 2, 3) e, (1, 2, 3)^{-1} (1, 2, 3) (1, 2, 3), (3, 2, 1)^{-1} (1, 2, 3) (3, 2, 1)\}= \{(1, 2, 3)\}.\)

\([(3, 2, 1)]=\{\sigma^{-1} (3, 2, 1) \sigma \mid \sigma \in A_3\}= \{e^{-1} (3, 2, 1) e, (1, 2, 3)^{-1} (3, 2, 1) (1, 2, 3), (3, 2, 1)^{-1} (3, 2, 1) (3, 2, 1)\}= \{(3, 2, 1) \}.\)

Since \(A_3 \) is abelian, conjugation by any element leaves every element unchanged.