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?
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 $H \subset G$.