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Math Help - result about finite groups

  1. #1
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    result about finite groups

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

    \\\mbox{Show that }\forall\,a,b\in G,\ ab\in H\ \Leftrightarrow\ \mbox{the orders}
    \\\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
    \\\mbox{Let $G$ be a finite group and $H$ a subgroup}
    \\\mbox{such that $|H|$ is odd and }|G:H|=2.

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

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