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Problem of the Week #223 - Sep 06, 2016

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Euge

MHB Global Moderator
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Jun 20, 2014
1,894
Here is this week's POTW:

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Let $G$ be a group with finite index subgroups $H$ and $K$. Suppose $H$ and $K$ have relatively prime indices in $G$. Why must $G$ be the internal direct product of $H$ and $K$?

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,894
No one answered this week's problem. You can read my solution below.


Since $(G : H)(H:H\cap K) = (G: H\cap K)= (G : K)(K:H\cap K)$ and $(G : H)$ is relatively prime to $(G : K)$, then $(G : H)$ divides $(K : H\cap K)$. Therefore, $(G : H) \le (K : H\cap K)$. On the other hand, the mapping $(H\cap K)k \overset{f}{\mapsto} kH$ from the right cosets of $H\cap K$ in $K$ to the right cosets of $H$ in $G$ is injective. Indeed, if $Hk = Hk'$, then $k(k')^{-1} = h$ for some $h\in H$. Since $K$ is a subgroup of $G$, then the equation $k(k')^{-1} = h$ implies $h\in K$ as well. So since $k = hk'$ and $h\in H\cap K$, then $(H\cap K)k = (H\cap K)k$. Thus, $(K : H\cap K) \le (G : H)$. Consequently, $(G : H) = (K : H\cap K)$ the mapping $f$ is surjective. Given $g\in G$, there exists a $k\in K$ such that $f(k) = Hg$, i.e., $(H\cap K)k = Hg$. Thus $xk = hg$ for some $x\in H\cap K$ and $h\in H$. Now $g = h^{-1}xk\in HK$ since $h^{-1}x\in H$ and $k\in K$. Since $g$ was arbitrary, we must have $G = HK$.
 
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