• Apr 10th 2009, 04:46 PM
siegfried
$\\\mbox{Let G be a finite group and H a subgroup}$
$\\\mbox{such that |H| is odd and }|G:H|=2.$

$\\\mbox{Show that }\forall\,a,b\in G,\ ab\in H\ \Leftrightarrow\ \mbox{the orders}$
$\\\mbox{of a and b are both odd or both even}.$
• Apr 10th 2009, 06:28 PM
ThePerfectHacker
Quote:

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

$\\\mbox{Show that }\forall\,a,b\in G,\ ab\in H\ \Leftrightarrow\ \mbox{the orders}$
$\\\mbox{of a and b are both odd or both even}.$

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 $ab\in H$.