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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 $G$?

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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 $\operatorname{Id}$.

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  • $\begingroup$ thanks, very easy $\endgroup$ – Domenico Vuono Dec 23 '18 at 13:07
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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.

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