# 3.2: Alternating Groups

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##### 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 $$A_4=\{e, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3), (1,3)(1,2), (1,4)(1,2),$$

$$(1,4)(1,3),(2,4)(2,3), (3,1)(3,2), (4,2)(4,1), (4,1)(4,3),(4,2)(4,3) \}$$.

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 $$A_4 \le A_n, \; \forall n \ge 4$$, $$A_n$$ 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.

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This page titled 3.2: Alternating Groups is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pamini Thangarajah.