To show a given set of matrices is a connected component of $GL_n(\mathbb{R})$ The set of $n\times n$ matrices can be identified with the space $\mathbb{R}^{n\times n}$ . Let $G$ be a subgroup of $GL_n(\mathbb{R})$. prove :
$(a)$ If $A$, $B$, $C$, $D$ are in $G$, and if there are paths in $G$ from $A$ to $B$ and from $C$ to $D$, then there is a path in $G$ from $AC$ to $BD$.
$(b)$ The set of matrices that can be joined to the identity $I$ form a normal subgroup of
$G$. (It is called the connected component of $G$.)
(A path from $A$ to $B$ is a continuous functions on $[0,1]$ with values in $\mathbb{R}^{n\times n}$, a function $X:[0,1]\mapsto \mathbb{R}^{n\times n}$ with $X(t)= (x_1(t),...x_k(t))$ such that $X(0)=A$ and $X(1) = B$)
How do I solve $(b)$?
Here's my approach: First I'm trying to prove that $N$ (given set of matrices in $(b)$) is a subgroup of $G$.
For $A, B \in N$ there exists continuous functions $X$ and $Y$ such that $X:[0,1]\mapsto\mathbb{R}^{n\times n}$ with $X(0)=I$ and $X(1)=A$ and $Y:[0,1]\mapsto\mathbb{R}^{n\times n}$ with $Y(0) = I$ and $Y(1) = B$ and we know that the product of two continuous functions is continuous, $(X.Y)$ will do the job, so  $AB\in N$. and for $A \in N$, we know that there exists a continuous function $Z:[0,1] \mapsto\mathbb{R}^{n\times n}$ with $Z(0) = I$ and $Z(1) = A$.
I think this might work for the inverse : $Z_1(t) = A^{-1}Z(t)$ but how do I show that this is continuous on [0,1]
How do I prove the inverse part? and also that N is a normal subgroup of  $GL_n(\mathbb{R})$?
 A: You can assert continuity by viewing the inverse $A^{-1}$ of a matrix $A$ on $\text{GL}_{n}$ in terms of its cofactors (as Tom kinsella mentions in the comments), and continuity of two matrices is also a continuous function.
So, for any two matrices $A, B \in G_{0}$ (where $G_{0}$ denotes the connected component of $\text{GL}_{n}$), there exist continuous paths from $I$ to $A$ and from $I$ to $B$, denoted by $A(t)$ and $B(t)$ respectively (as both $A$ and $B$ belong to the connected component $G_{0}$).
Then from the first paragraph, as the matrix multiplication of any two matrices is continuous, the path $A(t)B(t)$ is a continuous path from $I$ to $AB$, so $AB \in G_{0}$. Similarly, any matrix $A \in \text{GL}_{n}$ is invertible, and so the path $(A(t))^{-1}$ from $I$ to $A^{-1}$ is continuous since the path $A(t)$ from $I$ to $A$ is continuous.
Normality of $G_{0}$: Proceeds similarly to the above discussion: Let $A \in G_{0}$ and $B \in \text{GL}_{n}$ (so only $A$ has a path connecting it to $I$). Then $BA(t)B^{-1}$ is a path from $BA(0)B^{-1} = BIB^{-1} = I$ to $BAB^{-1}$, and thus $BAB^{-1} \in G_{0}$ as there is a continuous path connecting $I$ to $BAB^{-1}$.
