Problem of the Week #84 - January 6th, 2014

[Chris L T521]

Here's the first Graduate POTW of 2014!

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Problem: Let $G$ be a path-connected matrix group, and let $H$ be a subgroup of $G$ that contains a nonempty open subset $U$ of $G$. Show that $H=G$.

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

 

[Chris L T521]

This week's problem was correctly answered by Opalg and Turgul. You can find Opalg's solution below.

Spoiler

This result applies in any topological group, not just matrix groups. Also, it only requires the group to be connected (not necessarily path-connected). Let $h\in U$. Then $V = h^{-1}U \subseteq H$, so $V$ is an open neighbourhood of the identity contained in $H$. If $k\in H$ then $kV$ is an open neighbourhood of $k$ in $H$. That shows that $H$ is an open subset of $G$. But then every coset of $H$ is also open, because if $g\in G$ then $gV$ is an open neighbourhood of $g$ contained in the same coset $gH$ as $g$. Therefore the union of all the cosets of $H$ (other than $H$ itself) is open. Hence $H$, being the complement of that union, is closed. Thus $H$ is both open and closed (and nonempty, since it contains the identity element). Since $G$ is connected it follows that $H=G.$