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Let $G$ be a subgroup of $GL(n,\mathbb R)$ which is also a $m$-surface for some $m$, how to find a no where vanishing tangent vector field for $G$ ?

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    Take a non-zero tangent vector at the identity and translate it using the left (or right) multiplication of $G$ to any other point. If $G$ is connected, you're done. If not, do the same for the other components. – 2017-01-28
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    Do what levap suggest but for GL, starting with. Vector tangent to G, and then restrict to G. – 2017-01-28
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    Why don't you guys write answers instead of answering the question in the comments? – 2017-01-28
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    @levap : I don't understand what you are saying .. how to even get a non-zero tangent vector at the identity ? – 2017-01-28
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    @MarianoSuárez-Álvarez : I can get an element of $T_A(GL(n,\mathbb R))$ by considering $\alpha :(-r,r) \to GL(n, \mathbb R)$ as $\alpha (t)=Ae^{tX}$ , where $X$ is any arbitrary fixed matrix , showing that $T_A(GL(n,\mathbb R))=M(n,\mathbb R)$ .. but what do you mean by "restrict" it ? How do I get an element of $T_A(G)$ for at least even $A=I$ from here ? – 2017-01-28

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Recall that for any Lie group $G$ that left multiplication $L_g: G \to G$ given by $L_g(h) = gh$ is a diffeomorphism (the inverse given by $L_{g^{-1}}$), which furthermore shows that $d(L_g)_h: T_hG \to T_{gh}G$ is of full rank for all $h \in G$ (meaning that $\ker d(L_g)_h = \{0\} \subset T_hG$). Notice that for any Lie subgroup $H \subseteq G$ there exists some nonzero tangent vector $X_e \in T_eH$, hence the corresponding tangent vector $X_h \equiv d(L_h)_e(X_e)$ is nonzero since the Jacobian of left-translation is non-singular. By defining the vector field $X$ pointwise by $X(p) = X_p$, we see that this is smooth since if $\gamma:(-\epsilon, \epsilon) \to H$ is a curve with $\gamma(0) = e$ and $\gamma'(0) = X_e$ we find that for $f \in C^\infty(H)$

$$ (Xf)(p) \;\; =\;\; d(L_p)_e(X_e)f(p) \;\; =\;\; X_e(f\circ L_p)(e) \;\; =\;\; \left . \frac{d}{dt} (f\circ L_g \circ \gamma) \right |_{t=0} $$

which is a smooth map. This proves moreover that $X$ is a smooth vector field that doesn't vanish anywhere.

We can see that we can simply apply the above argument to $GL(n,\mathbb{R})$ and to any appropriate Lie subgroup.

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    Notice that (using your notations) the OP wants $H$ to be a subgroup and a submanifold of $G$, but not necessarily a Lie subgroup. At least this is what the question asks - whether this is indeed what the OP meant is another discussion. – 2017-01-31
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    @AlexM. You have a point there, however I don't see how a subgroup which is also an "$m$-surface" cannot be a Lie subgroup. Assuming the subgroup is a surface then multiplication and inversion are smooth by restricting the operations on $GL(n,\mathbb{R})$ to the subgroup in question. – 2017-01-31