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

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#### Chris L T521

##### Well-known member
Staff member
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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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 $H=G.$