# Normal subgroup in a matrix Lie group

Prop: If $$G$$ is a Matrix Lie group, then the connected component that contains the identity $$I$$ is a normal subgroup of $$G$$.

I have problem in the proof of this. Suppose that $$A$$ and $$B$$ belong to the connected component that contains $$I$$, then exist two continuos function ($$A(t), B(t)$$) in $$G$$ such that $$A(0)=B(0)=I$$ and $$A(1)=A$$, $$B(1)=B$$.

My question is: if we consider $$A(t)B(t)$$ why this path is contained in the connected component of $$G$$?

Because the mapt $$t\mapsto A(t)B(t)$$ is continuous and its domain is conneced (it is an interval). Therefore, its range is connected too. Since, furtheremore, $$\operatorname{Id}$$ belongs to the range, the range is a subset of the connected component of $$\operatorname{Id}$$.
If $$\gamma\colon[0,1]$$ is a path from $$a$$ to $$b$$, then $$a^{-1}\gamma$$ is a path from $$1$$ to $$a^{-1}b$$. Hence in any topological group, the path-connected component of $$1$$ is a subgroup.
Moreover, for $$g\in G$$, if $$\gamma$$ is a path from $$1$$ to $$a$$, then $$g^{-1}\gamma g$$ is a path from $$1$$ to $$g^{-1}ag$$, hence that subroup is normal.