TITLE: Showing path connected matrices of a group $G$ is a normal subgroup
QUESTION [6 upvotes]: Let $G$ be a subgroup of $GL_n(\Bbb{R})$. Define $$H = \biggl\{ A \in G  \ \biggl| \ \exists \ \varphi:[0,1] \to G \ \text{continuous such that} \ \varphi(0)=A , \ \varphi(1)=I\biggr\}$$ Show that $H$ is a normal subgroup of $G$
In this post, the answers are either too advanced (manifolds) or are considered overkill. I was trying to follow the user jug's hints, but I can't understand how to make use of  $\phi_A(x) = AxA^{-1}$. I believe I have the same confusion as the OP -- I want to show that for all $A \in G$ and for all $M \in H$, $AMA^{-1} \in H$ -- but I'm struggling very much connecting the idea of a normal subgroup with a continuous function (or path).
Could someone please expand on the hints provided in the linked post?

REPLY [7 votes]: Let $H$ be the path-connected component of $I$ in $G$.

First, let us notice that $H$ is not empty.
Suppose that $A$ and $B$ are in $H$, so that there are maps $\sigma:I\to G$ and $\tau:I\to G$ such that $\sigma(0)=I$, $\sigma(1)=A$, $\tau(0)=I$ and $\tau(1)=B$. Consider the map $$\lambda:t\in I\longmapsto \sigma(t)\tau(t)^{-1}\in G.$$ You can show that 

the map $\lambda$ is continuous, and that
$\lambda(0)=I$ and $\lambda(1)=AB^{-1}$.

If follows from this that $AB^{-1}$ is also in $H$.
Suppose $A$ is in the path-connected component $H$ of $I$, so that there is a path $\sigma:I\to G$ such that $\sigma(0)=I$ and $\sigma(1)=A$.
Let now $B\in G$ and consider the map $$\tau:t\in[0,1]\longmapsto B\sigma(t)B^{-1}\in G.$$ You should check that

it is continuous,
its value at $0$ is $I$, and its value at $1$ is $BAB^{-1}$.

It follows from all this that $BAB^{-1}$ is also an element of $H$.

Now conclude that $H$ is a normal subgroup.