Connected component of $GL_n(\mathbb{R})$ The set of $n\times n$ matrices can be identified with the space $M_n(\mathbb{R})$. Let $G$ be a subgroup of $GL_n(\mathbb{R})$.
I would like to show that the set of matrices that can be joined to the identity $I$ forms a normal subgroup of G.
 A: If $\gamma:[0,1]\to G$ is a path, $\gamma(0)=g$, $\gamma(1)=1$, then $\gamma'(t):=h\gamma(t)h^{-1}$ is a path from $hgh^{-1}$ to $1$. Your set is thus invariant wrt. conjugations. To see that it is a subgroup, if $\gamma_1$ and $\gamma _2$ are two paths, consider the new path $\gamma(t):=\gamma_1(t)\gamma_2(t)$ (so it's closed under product), or $\gamma(t)=\gamma_1(t)^{-1}$ (closed under inverse).
A: This is true for any topological group $G$ : The basic idea is that the two maps
$$
R_g : x \mapsto xg \text{ and } L_g : x\mapsto gx
$$
are both homeomorphisms.
Let $K$ be the connected component of the identity, then for any $x \in K$,
$$
Kx^{-1} = R_{x^{-1}}(K)
$$
is connected (since it is the image of $K$ under a homeomorphism) and contains $e=xx^{-1}$. Hence,
$$
Kx^{-1} \subset K \Rightarrow KK^{-1} \subset K
$$
Hence, $K$ is a subgroup.
Furthermore, if $g\in G$, then
$$
gKg^{-1} = R_{g^{-1}}L_g(K)
$$
is connected and contains the identity. Hence $gKg^{-1} \subset K$, and hence $K$ is normal.
