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Thread: Normal subgroups

  1. #1
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    [SOLVED] Normal subgroups

    we know that if a group is abelian, all subgroups of it are normal. however - can you prove or disprove that if all subgroups of a group are normal, it is abelian?
    Last edited by shos; Aug 24th 2010 at 10:47 PM.
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  2. #2
    Lord of certain Rings
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    We will disprove the claim.

    Consider the group $\displaystyle Q_8 = \{\pm1,\pm i,\pm j,\pm k\}$ where $\displaystyle i^2 = j^2 = k^2 = -1, (-1)^2 = 1$. Also we need $\displaystyle ijk = -1$. Using the above relations you can show that $\displaystyle Q_8$ is a group. You can also show it is non-abelian by observing that $\displaystyle ij = -ji$.
    Note that $\displaystyle Q_8$ is generated by any two of $\displaystyle i,j,k$.

    Now show that if any subgroup $\displaystyle H$ contains $\displaystyle i$ then $\displaystyle \{\pm 1, \pm i\} \subseteq H$. If $\displaystyle H = \{\pm 1, \pm i\}$, then index of $\displaystyle H$ in $\displaystyle Q_8$ is 2 and thus $\displaystyle H$ is normal in $\displaystyle Q_8$. Suppose $\displaystyle \{\pm 1, \pm i\} \subsetneq H$, then one of $\displaystyle j$ or $\displaystyle k$ must belong to $\displaystyle H$. But then $\displaystyle H = Q_8$.

    We can proceed similarly for the case where a subgroup $\displaystyle H$ contains $\displaystyle j$ or $\displaystyle k$.
    The only remaining case is that $\displaystyle H$ does not contain $\displaystyle i$,$\displaystyle j$ or $\displaystyle k$. In this case $\displaystyle H = \{1,-1\}$, which is the center of $\displaystyle Q_8$ and thus $\displaystyle H$ is normal in $\displaystyle Q_8$
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