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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