In this post, Terry Tao says that Gaussians are "a subgroup of $\Bbb R$".
When you convolve Haar probability measure on a (compact) subgroup with itself, you get back the same measure, and this can in fact be used as a definition of such subgroups. If you convolve a Gaussian probability measure with itself, you almost get back the same Gaussian measure, but it has spread out by a factor of $\sqrt 2$. So Gaussians are in some sense a "$\sqrt 2$-approximate group".
I don't understand this comment, or how it means that Gaussians can be viewed as a subgroup of $\Bbb R$.
How can you define subgroups through this? Perhaps this?
A subset, $H$ of a group $G$ with Haar measure $\mu$ is called a subgroup if $\mu |_H=\mu\ast\mu|_H$.
How does this show that a Gaussian is a subgroup of $\Bbb R$?
Obviously, I am not Terry Tao and this answer has some speculation on my part. The interesting bit of theory is that subgroups (of a compact group) can be identified by algebraic properties in the semigroup of probability measure. This post only aims to answer the question, "How can you define a subgroup through this?".
Let $G$ be a compact group and $\mathcal{P}(G)$ be the collection of regular Borel probability measures on $G$. This is a semigroup under the convolution operation. There is a classical theorem of Wendel which connects idempotent elements in the semigroup $\mathcal{P}(G)$ (i.e. measures such that $\mu * \mu = \mu$) with compact subgroups of $G$. In particular, each idempotent measure is actually the Haar measure of a closed subgroup. So, from an idempotent measure, taking the support gives a compact subgroup, and from a compact subgroup, taking the Haar measure results in an idempotent measure.
So, we can say that a subset $A \subset G$ is a (compact) subgroup if there exists an idempotent measure $\mu \in \mathcal{P}(G)$ such that $\sup(\mu) = A$.
Rudin extended this theorem to the locally compact abelian case. In particular, any idempotent probability measure here concentrates on a compact subgroup.