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Thread: result about finite groups

  1. #1
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    result about finite groups

    $\displaystyle \\\mbox{Let $G$ be a finite group and $H$ a subgroup}$
    $\displaystyle \\\mbox{such that $|H|$ is odd and }|G:H|=2.$

    $\displaystyle \\\mbox{Show that }\forall\,a,b\in G,\ ab\in H\ \Leftrightarrow\ \mbox{the orders}$
    $\displaystyle \\\mbox{of $a$ and $b$ are both odd or both even}.$
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  2. #2
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    Quote Originally Posted by siegfried View Post
    $\displaystyle \\\mbox{Let $G$ be a finite group and $H$ a subgroup}$
    $\displaystyle \\\mbox{such that $|H|$ is odd and }|G:H|=2.$

    $\displaystyle \\\mbox{Show that }\forall\,a,b\in G,\ ab\in H\ \Leftrightarrow\ \mbox{the orders}$
    $\displaystyle \\\mbox{of $a$ and $b$ are both odd or both even}.$
    Here is the forward direction: if $\displaystyle ab\in H$ then $\displaystyle |a|\equiv |b|(\bmod 2)$. If $\displaystyle a\in H$ then $\displaystyle (a^{-1})(ab) = b \in H$, so $\displaystyle |a|\text{ and }|b|$ will then be both odd since $\displaystyle |H|$ is odd. If $\displaystyle b\in H$ then by a similar argument $\displaystyle a\in H$, so $\displaystyle |a|\text{ and }|b|$ will then both be odd. Therefore, it is safe to assume that $\displaystyle a\not \in H\text{ and }b\not \in H$. Since $\displaystyle (G:H)=2$ it means $\displaystyle H$ is a normal subgroup and we can form the factor group $\displaystyle G/H$. Notice that (in the case we are assuming) $\displaystyle aH=bH\not = e$. Therefore, the order of $\displaystyle aH$ and $\displaystyle bH$ in $\displaystyle G/H$ is $\displaystyle 2$. Let $\displaystyle |a|=i,|b|=j$. Then $\displaystyle (aH)^i = a^i H = H\text{ and }(bH)^j = b^j H = H$. Thus, $\displaystyle 2|i \text{ and }2|j$, hence both are even. Thus, $\displaystyle |a|,|b|$ always have the same parity.

    Now for the converse direction. We will work in cases, first we will assume that $\displaystyle |a|,|b|$ are both odd and then that both are even. If both are odd then order of $\displaystyle aH$ in $\displaystyle G/H$ cannot be $\displaystyle 2$ (by above argument) so $\displaystyle aH=H$, likewise $\displaystyle bH=H$. If both are even then $\displaystyle a,b\not \in H$ since $\displaystyle |H|$ is odd and so $\displaystyle aH=bH$ (since there are only two cosets the ones different from $\displaystyle H$ must be equal). Regardless, in whatever case $\displaystyle aH=bH$ and so $\displaystyle abH = (aH)(bH) = H$, so $\displaystyle ab\in H$.
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