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Thread: 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 $\displaystyle K=G\times H,$ where $\displaystyle G$ is an abelian group and $\displaystyle H$ is a cyclic group of order 2, and suppose $\displaystyle G$ contains an element $\displaystyle g$ with $\displaystyle g^2\ne1_G.$ Find a binary operation $\displaystyle \circ$ on $\displaystyle K$ such that $\displaystyle \left<K,\circ\right>$ 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 $\displaystyle K=G\times H,$ where $\displaystyle G$ is an abelian group and $\displaystyle H$ is a cyclic group of order 2, and suppose $\displaystyle G$ contains an element $\displaystyle g$ with $\displaystyle g^2\ne1_G.$ Find a binary operation $\displaystyle \circ$ on $\displaystyle K$ such that $\displaystyle \left<K,\circ\right>$ 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 $\displaystyle H=\{1, x \}.$ define $\displaystyle \theta: H \longrightarrow \text{Aut}(G)$ by $\displaystyle \theta(1)=\text{id}_G$ and $\displaystyle \theta(x)(a)=a^{-1}, \ \forall a \in G.$ now define $\displaystyle \circ$ by $\displaystyle (g_1,h_1) \circ (g_2,h_2)=(g_1 \theta(h_1)(g_2), h_1h_2), \ \ \forall g_1,g_2 \in G$ and $\displaystyle \forall h_1,h_2 \in H.$

    then $\displaystyle <K, \circ>$ is a group because $\displaystyle <K, \circ> \cong G \rtimes_{\theta} H$ and it's non-abelian because for any $\displaystyle g \in G$ with $\displaystyle g^2 \neq 1$ we have: $\displaystyle (1,x) \circ (g,1) \neq (g,1) \circ (1,x).$
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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 $\displaystyle H=\{1, x \}.$ define $\displaystyle \theta: H \longrightarrow \text{Aut}(G)$ by $\displaystyle \theta(1)=\text{id}_G$ and $\displaystyle \theta(x)(a)=a^{-1}, \ \forall a \in G.$ now define $\displaystyle \circ$ by $\displaystyle (g_1,h_1) \circ (g_2,h_2)=(g_1 \theta(h_1)(g_2), h_1h_2), \ \ \forall g_1,g_2 \in G$ and $\displaystyle \forall h_1,h_2 \in H.$

    then $\displaystyle <K, \circ>$ is a group because $\displaystyle <K, \circ> \cong G \rtimes_{\theta} H$ and it's non-abelian because for any $\displaystyle g \in G$ with $\displaystyle g^2 \neq 1$ we have: $\displaystyle (1,x) \circ (g,1) \neq (g,1) \circ (1,x).$
    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 $\displaystyle g_1,g_2\in G,\ H=\left<h\right>,$

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

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

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

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


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

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

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

    $\displaystyle K$ is known as the cyclic extension of $\displaystyle G$ by $\displaystyle H$ induced by the automorphism $\displaystyle \vartheta.$ 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, $\displaystyle \vartheta$ is the mapping $\displaystyle g\mapsto g^{-1}$ which is an automorphism because $\displaystyle G$ is Abelian. And $\displaystyle K$ is not Abelian because $\displaystyle (g,1)(g,h)=(g^2,h)$ whereas $\displaystyle (g,h)(g,1)=(1,h);$ the two are not equal if $\displaystyle g^2\ne1.$
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