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Thread: n on-normal subgroup of a direct product group

  1. #1
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    n on-normal subgroup of a direct product group

    This one has kicked my *ss. I am trying to find a subgroup of $\displaystyle Q_8 \times Z_4$ that is not normal.

    Now, all subgroups of $\displaystyle Q_8$ are normal and all subgroups of $\displaystyle Z_4$ are normal. Then I need to find a subgroup of $\displaystyle Q_8 \times Z_4$ that is not normal.

    The only case that is nontrivial is the subgroup $\displaystyle < i > \times (1, b)$. ( (1, b) where $\displaystyle b^2 = 1$ is the only subgroup of $\displaystyle Z_4$.)

    Notation: I am defining
    $\displaystyle (a, c) \cdot < i > \times (1, b) = ( a < i > ) \times (c, cb)$ where a is an arbitrary element of $\displaystyle Q_8$ and c is an arbitrary element of $\displaystyle Z_4$.

    and
    $\displaystyle < i > \times (1, b) \cdot (a, c) = ( < i > a ) \times (c, bc)$

    Okay, so given an arbitrary a and c we have
    $\displaystyle (a, c) \cdot < i > \times (1, b) = ( a < i > ) \times ( c, cb) $ $\displaystyle = (a < i > ) \times (c, bc )$
    because $\displaystyle Z_4$ is Abelian.

    Since $\displaystyle < i >$ is normal in $\displaystyle Q_8$ the above is equal to
    $\displaystyle (< i > a ) \times (c, bc) = < i >\times (1, b) \cdot (a, c) = ( a, c ) \cdot < i > \times ( 1, b )$.

    So $\displaystyle < i > \times (1, b)$ is normal in $\displaystyle Q_8 \times Z_4$ for all a, c.

    Where have I gone wrong?

    Thanks!

    -Dan
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    Re: n on-normal subgroup of a direct product group

    Let $a$ be the generator of $Z_4$. Consider $H = <(i,a)>$.

    You have $H = \{(1,1),(i,a),(-1,a^2),(-i,a^3)\}$

    Now, $(j,a)(i,a)(-j,a^3) = (-i,a) \notin H$.

    I think where you went wrong was assuming you needed to take a product of subgroups rather than finding the smallest subgroup containing an element of the product.
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    Forum Admin topsquark's Avatar
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    Re: n on-normal subgroup of a direct product group

    Quote Originally Posted by SlipEternal View Post
    Let $a$ be the generator of $Z_4$. Consider $H = <(i,a)>$.

    You have $H = \{(1,1),(i,a),(-1,a^2),(-i,a^3)\}$

    Now, $(j,a)(i,a)(-j,a^3) = (-i,a) \notin H$.

    I think where you went wrong was assuming you needed to take a product of subgroups rather than finding the smallest subgroup containing an element of the product.
    I have a nasty tendency to do things the hard way. Very nice.

    Thanks!

    -Dan
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