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Math Help - TAs Fun Algebra Problem

  1. #1
    Senior Member TheAbstractionist's Avatar
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    TAs Fun Algebra Problem

    Hi all.

    Following NonCommAlgs example, I have decided to post a fun algebra (group theory) problem of my own on this forum.

    Let K=G\times H, where G is an abelian group and H is a cyclic group of order 2, and suppose G contains an element g with g^2\ne1_G. Find a binary operation \circ on K such that \left<K,\circ\right> is a nonabelian group.

    To NonCommAlg: If you know the answer to this one (and I have a feeling you do) please dont be too quick to reveal the answer. Let others have a chance to do the problem first.
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  2. #2
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    Quote Originally Posted by TheAbstractionist View Post
    Hi all.

    Following NonCommAlgs example, I have decided to post a fun algebra (group theory) problem of my own on this forum.
    Let K=G\times H, where G is an abelian group and H is a cyclic group of order 2, and suppose G contains an element g with g^2\ne1_G. Find a binary operation \circ on K such that \left<K,\circ\right> is a nonabelian group.
    To NonCommAlg: If you know the answer to this one (and I have a feeling you do) please dont be too quick to reveal the answer. Let others have a chance to do the problem first.
    it took me some time to figure this one out!


    Spoiler:
    let H=\{1, x \}. define \theta: H \longrightarrow \text{Aut}(G) by \theta(1)=\text{id}_G and \theta(x)(a)=a^{-1}, \ \forall a \in G. now define \circ by (g_1,h_1) \circ (g_2,h_2)=(g_1 \theta(h_1)(g_2), h_1h_2), \ \ \forall g_1,g_2 \in G and  \forall h_1,h_2 \in H.

    then <K, \circ> is a group because <K, \circ> \cong G \rtimes_{\theta} H and it's non-abelian because for any g \in G with g^2 \neq 1 we have: (1,x) \circ (g,1) \neq (g,1) \circ (1,x).
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    it took me some time to figure this one out!


    Spoiler:
    let H=\{1, x \}. define \theta: H \longrightarrow \text{Aut}(G) by \theta(1)=\text{id}_G and \theta(x)(a)=a^{-1}, \ \forall a \in G. now define \circ by (g_1,h_1) \circ (g_2,h_2)=(g_1 \theta(h_1)(g_2), h_1h_2), \ \ \forall g_1,g_2 \in G and  \forall h_1,h_2 \in H.

    then <K, \circ> is a group because <K, \circ> \cong G \rtimes_{\theta} H and it's non-abelian because for any g \in G with g^2 \neq 1 we have: (1,x) \circ (g,1) \neq (g,1) \circ (1,x).
    Hi NonCommAlg.

    Yes, that was what I had in mind.

    For those who don’t understand NonCommAlg’s proof:
    Spoiler:
    The binary operation is this: For g_1,g_2\in G,\ H=\left<h\right>,

    \left(g_1,1_H\right)\circ\left(g_2,1_H\right)\ =\ \left(g_1g_2,1_H\right)

    \left(g_1,1_H\right)\circ\left(g_2,h\right)\ =\ \left(g_1g_2,h\right)

    \left(g_1,h\right)\circ\left(g_2,1_H\right)\ =\ \left(g_1g_2^{-1},h\right)

    \left(g_1,h\right)\circ\left(g_2,h\right)\ =\ \left(g_1g_2^{-1},1_H\right)


    Comments:
    Spoiler:
    You can verify that all the group axioms are satisfied for the operation defined above, but checking associativity directly can be rather time consuming. Fortunately, we have a theorem.

    Let G be any group, H=\left<h\right> a cyclic group of order r and \vartheta an automorphism of G such that \vartheta^r is the identity automorphism. Then K=G\times H is a group with respect to the operation \circ where

    \left(g_1,h^i\right)\circ\left(g_2,h^j\right)\ =\ \left(g_1\vartheta^i(g_2),h^{i+j}\right)

    for all g_1,g_2\in G and i,j\in\mathbb Z.

    K is known as the cyclic extension of G by H induced by the automorphism \vartheta. For a proof of the theorem (and more on cyclic extensions) see Chapter 21 of John F. Humphreys, A Course in Group Theory, Oxford University Press, 1996.

    In our example, \vartheta is the mapping g\mapsto g^{-1} which is an automorphism because G is Abelian. And K is not Abelian because (g,1)(g,h)=(g^2,h) whereas (g,h)(g,1)=(1,h); the two are not equal if g^2\ne1.
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