Do closed subgroups of connected groups have open connected components? Let $G$ be a topological group which is connected and let $H$ be a subgroup.
Is it true that $H$ is closed in $G$ if and only if the connected component of the identity in $H$ (which has the subspace topology) is open (as a subset of $H$)?
If not what are the counterexamples?
In the case I care about $G$ is a simply connected Lie group; while it seems to me that the answer to the above question should be yes in general, is the answer yes at least under this stronger assumption?
 A: The 'if' direction is false. Say $G=T^2$ is the 2-torus and $H$ is an irrationally sloped line in $T^2$. Then $H$ is connected as it's the continuous image of a connected set $\mathbb{R}$. Hence, the connected component of the identity in $H$ is $H$, which is open in $H$. However, $H$ is not closed in $T^2$. In fact, $H$ is a proper dense subset of $T^2$. This example generalizes to any Lie group that contains a 2-torus. In particular, $SU(n)$ for $n \ge 3$ is simply-connected and contains a 2-torus, so the 'if' direction is false even if $G$ is a simply-connected Lie group.
On the other hand, if $H$ is a closed subgroup of a Lie group $G$ then any connected component in $H$ is open. To see why, let $A$ be a connected component $A \subset H$ and $x \in A$ an arbitrary point in $A$. The exponential map can be used to define a diffeomorphism from a connected open subset of the Lie algebra $\mathfrak{h}$ of $H$ to a connected open subset of $H$ containing $x$. This connected open subset of $H$ containing $x$ must lie in the connected component $A$. Since $x \in A$ was arbitrary, it follows that $A$ is open. Note that since $H \subset G$ is closed in this case, the group topology on $H$ is the same as the relative topology of $H \subset G$.
