In Andrew Baker's book 'Matrix Groups', he as defined tangent space to $G$, a matrix group, at $U \in G$ as $$T_U G =\{\gamma ' (0) \in M_n(\mathbb K) : \gamma \ \text{is differentiable curve in} \ G \ \text{with} \ \gamma(0) =U \} $$ In proving that $T_UG$ is real vector space to show closure under addition he uses following construction $\gamma (t)=\alpha(t) U^{-1} \beta(t) $. Thus it is clear that this proof depends on $U$ being invertible.
But in general tangent space of a manifold is defined using pushforward. The two definitions are equivalent. So why invertibility of $U$ seems to play important role here? Is there alternate proof other than the argument that the two definitions of tangent space are equivalent and thus...

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    $\begingroup$ In general, you use smoothness of $M$. In the case of matrix group, it is the group structure that gurantees smoothness (i.e., if it is possible to move infiniteismally into two directions, then by the group property, it is possible to move infinitesimally into the sum direction). Alternatively, you could restrict yourself to groups that have a sufficiently nice definition (e.g., algebraic groups, i.e., given by polynomial equations in the matrix entries) $\endgroup$ – Hagen von Eitzen May 19 at 11:45

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