Let $M(n,\mathbb{R}$) denote the set of all $n \times n$ matrices with real entries (identified with $\mathbb{R}^{n^2}$ and endowed with its usual topology) and let $GL(n, \mathbb{R})$ denote the group of invertible matrices. Let $G$ be a subgroup of $GL(n, \mathbb{R})$. Define $$H = \begin{cases} A \in G \ |\ \text{there exists}\ \phi : [0, 1] \to G \ \text{continuous},\\ \hspace{20mm} \text{such that}\ \phi(0) = A, \phi(1) = I \end{cases}$$ Then, $H$ is a normal subgroup of $G$.
My starting- Let $A \in H$ and $B \in G$, then to prove $ABA^{-1} \in H$. As $A \in H$ means there exists $\phi : [0, 1] \to G$ continuous and $\phi(0) = A$ and $\phi(1) = I$. Any idea how to go ahead?
That's not what being normal means. What you need to prove is that if $A\in H$ and $B\in G$, then $BAB^{-1}\in H$. Take $\phi\colon[0,1]\longrightarrow G$ continuous such that $\phi(0)=A$ and $\phi(1)=\operatorname{Id}$. Now, define $\phi_B(t)=B\phi(t)B^{-1}$. So, $\phi_B(0)=BAB^{-1}$ and $\phi_B(1)=B\operatorname{Id}B^{-1}=\operatorname{Id}$. Besides, $\phi_B$ is continuous.