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Well, the crux of the matter is to cook up the linear subspace of $gl(n)$ that will eventually be the subalgebra.

If $H$ is the subgroup: set $h={X\in gl(n): (\forall t \in \mathbb{R}) \, exp(tX) \in H }$.

You need to show two not completely obvious things:

1. $h$ is a linear subspace

2. $exp(h)$ is a nbhd of $1$ in $H$ (give $H$ the induced topology from $GL_n(\mathbb{R})$).

Adams has ingenious short arguments for these on pages 17-19 of his Lectures on Lie groups which require nothing but a little real analysis.

I remark that all this uses no manifold or Lie theory beyond the necessary definitions and the argument works completely without change for closed subgroups of any Lie group.

Well, the crux of the matter is to cook up the linear subspace of $gl(n)$ that will eventually be the subalgebra.

If $H$ is the subgroup: set $h=\{X\in gl(n): (\forall t \in \mathbb{R}) \, exp(tX) \in H \}$.

You need to show two not completely obvious things:

1. $h$ is a linear subspace

2. $exp(h)$ is a nbhd of $1$ in $H$ (give $H$ the induced topology from $GL_n(\mathbb{R})$).

Adams has ingenious short arguments for these on pages 17-19 of his Lectures on Lie groups which require nothing but a little real analysis.

I remark that all this uses no manifold or Lie theory beyond the necessary definitions and the argument works completely without change for closed subgroups of any Lie group.

Well, the crux of the matter is to cook up the linear subspace of gl(n) that will eventually be the subalgebra.

If H is the subgroup: set $h={X\in gl(n): (\forall t \in \mathbb{R}) \, exp(tX) \in H }$.

You need to show two not completely obvious things:

1. h is a linear subspace

2. $exp(h)$ is a nbhd of 1 in H (give H the induced topology from $GL_n(\mathbb{R})$).

Adams has ingenious short arguments for these on pages 17-19 of his Lectures on Lie groups which require nothing but a little real analysis.

I remark that all this uses no manifold or Lie theory beyond the necessary definitions and the argument works completely without change for closed subgroups of any Lie group.

Well, the crux of the matter is to cook up the linear subspace of $gl(n)$ that will eventually be the subalgebra.

If $H$ is the subgroup: set $h={X\in gl(n): (\forall t \in \mathbb{R}) \, exp(tX) \in H }$.

You need to show two not completely obvious things:

1. $h$ is a linear subspace

2. $exp(h)$ is a nbhd of $1$ in $H$ (give $H$ the induced topology from $GL_n(\mathbb{R})$).

Adams has ingenious short arguments for these on pages 17-19 of his Lectures on Lie groups which require nothing but a little real analysis.

I remark that all this uses no manifold or Lie theory beyond the necessary definitions and the argument works completely without change for closed subgroups of any Lie group.

Well, the crux of the matter is to cook up the linear subspace of gl(n) that will eventually be the subalgebra.

If H is the subgroup: set h={X\in gl(n): exp(tX) \in H for all t \in\R}.

You need to show two not completely obvious things:

1. h is a linear subspace

2. exp(h) is a nbhd of 1 in H (give H the induced topology from GL(n)).

Adams has ingenious short arguments for these on pages 17-19 of his Lectures on Lie groups which require nothing but a little real analysis.

I remark that all this uses no manifold or Lie theory beyond the necessary definitions and the argument works completely without change for closed subgroups of any Lie group.

Well, the crux of the matter is to cook up the linear subspace of gl(n) that will eventually be the subalgebra.

If H is the subgroup: set $h={X\in gl(n): (\forall t \in \mathbb{R}) \, exp(tX) \in H }$.

You need to show two not completely obvious things:

1. h is a linear subspace

2. $exp(h)$ is a nbhd of 1 in H (give H the induced topology from $GL_n(\mathbb{R})$).

Adams has ingenious short arguments for these on pages 17-19 of his Lectures on Lie groups which require nothing but a little real analysis.

I remark that all this uses no manifold or Lie theory beyond the necessary definitions and the argument works completely without change for closed subgroups of any Lie group.