3.2: Alternating Groups

Definition: Alternating groups

Alternating groups $A_n$ is the set of all even permutations associated with composition.  $|A_n|=\frac{n!}{2}$. $A_n$ is a subgroup of the symmetric group $S_n.$

Example $\PageIndex{1}$

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

Theorem $\PageIndex{1}$

$A_n$  is non abelian for $n \ge 4$.

Proof:

Let $a,b \in A_4$.

We will show that $ab \ne ba, \; \forall a,b \in A_n$, $n \ge 4$.

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 $ab=(1,2)(3,4)(4,1)(4,2)=(1,3,4,2)$.

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

Since $(1,3,4,2) \ne (1,4,3,2), \; \forall a,b \in A_4$, thus $A_4$ is non-abelian.

Since a copy of $A_4$ is a subgroup of $A_n, \; \forall n \ge 4$, $A_n$ is non-abelian for $N\ge 4$.◻

Example $\PageIndex{1}$

Find a cyclic subgroup of $A_8$ that has order 4.  

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

Then consider $(1,2,3,4)(1,2,3,4)=(1,3)(2,4) \in A_8$.

Further $(1,2,3,4)(1,3)(2,4)=(1,4,3,2) \in A_8$.

Further $(1,2,3,4)(1,4,3,2)=e$.

Thus $\{e, (1,2,3,4),(1,3)(2,4),(4,3,2,1)\}$, a subgroup of $A_8$ that is order 4 and cyclic.

 

Example $\PageIndex{2}$

 Find a non-cyclic subgroup of $A_8$ that has order 4.

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

Then consider $(1,2)^n$.

There is no $n$ for which $(1,2)^n=(3,4) \text{ or } (2,3)$.

Similarly, there is no $n$ for which $(2,3)^n=(1,2)\text{ or } (3,4)$.

Similarly, there is no $n$ for which $(3,4)^n=(1,2)\text{ or } (2,3)$.

Thus $\{e,(1,2),(2,3),(3,4)\}$ is not cyclic.

Version 2:

Let $G=\{e,i,j,k\}$.

Let $i=(1,2)(3,4)$, $j=(5,6)(7,8)$ and $k=(1,2)(3,4)(5,6)(7,8)$.

Consider $ij=k$, $ik=j$, and $jk=i$.

Thus there is closure in $G$.

Further, consider $i^{-1}=i$, $j^{-1}=j$, and $k^{-1}=k$.

Thus, inverses exist for all elements of $G$.

Note there is no $n$ for which $i^n=j$ or $k$.

Similarly, there is no $n$ for which $j^n=i$ or $k$.

Similarly, there is no $n$ for which $k^n=j$ or $k$.

Since there is no $<g> \in G$, $G$ is not cyclic.

Example $\PageIndex{3}$

Find the conjugacy classes of $A_3$.

Solution

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

$[e]=\{\sigma^{-1} e\sigma \mid \sigma \in A_4\}=\{e\}.$

$[(1, 2, 3)]=\{\sigma^{-1} (1, 2, 3) \sigma \mid \sigma \in A_4\}= \{e^{-1} (1, 2, 3) e, (1, 2, 3)^{-1} (1, 2, 3) (1, 2, 3), (3, 2, 1)^{-1} (1, 2, 3) (3, 2, 1)\}= \{(1, 2, 3)\}.$

$[(3, 2, 1)]=\{\sigma^{-1} (3, 2, 1) \sigma \mid \sigma \in A_4\}= \{e^{-1} (3, 2, 1) e, (1, 2, 3)^{-1} (3, 2, 1) (1, 2, 3), (3, 2, 1)^{-1} (3, 2, 1) (3, 2, 1)\}= \{(3, 2, 1) \}.$