Suppose $\phi:H \looparrowright G$ is an immersion of Lie groups. I'd like to establish the following claim:
If $\gamma:\mathbb{R} \to G$ is a smooth path with $\gamma(0)=e$ and $\gamma(t) \in \phi(H)$ for all $t$, then $\gamma'(0) \in d\phi(\operatorname{Lie} H)$.
The worry here is there on the level of manifolds, there are immersions like the map $\mathbb{R} \to \mathbb{R}^2$ which takes the line and bends the ends back around to limit towards the same point in the interior, like a figure eight:
Here the vertical curve through the origin lies in the image of the immersion but it's not "in the tangent space". The thought is that such a thing cannot happen in Lie groups, because we can translate such a singularity about by multiplication and find it everywhere. This in turn seems unlikely to be possible. I'd expect that to follow from Sard's theorem, but haven't been able to quite figure out an argument.
**This came up in trying to show that normal immersed subgroups of a Lie group give ideals of the Lie algebra. You'd like to show that conjugation by $G$ preserves $d\phi(\operatorname{Lie}H)$ because conjugation preserves $H$, but you'd need to outlaw an example as above from occurring.