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3.2: Alternating Groups

  • Page ID
    131874
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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 \(A_4=\{e, (12)(34), (13)(24), (14)(23), (123), (124), (134), (234), (432), (431), (421), (321) \}\). 

    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

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

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

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

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

     


    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.

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