a question on normal subgroup of $GL_n(\mathbb{C})$ and $GL_n(\mathbb{R})$ 
I am really sorry that I am not able to solve this one, thank you for your help.
 A: $a$ and $c$ are true by abstract group theory.
The first one shoudn't be too hard. Just write it out.
The third one basically says that the path component of identity in a topological group is a normal subgroup. To show it, for arbitrary $\varphi:[0,1]\to G$, $a\in G$, consider $\varphi_a(x)=axa^{-1}$.
The second one isn't true. To see this notice that conjugation in general linear group is just a change of basis, so \begin{pmatrix}1 &1\\ 0 &1\end{pmatrix} and \begin{pmatrix}1 &0\\ 1 &1\end{pmatrix} are clearly conjugate.
Edit:
I just noticed that the hint I've dropped for c only shows how to see that $H$ is normal, not that it is actually a subgroup. For that, take $a,b\in H$ corresponding to paths $\varphi,\psi$ and consider $\varphi\cdot \psi$ (pointwise multiplication) and for each $a\in H$ consider $\varphi^{-1}$ (not the inverse function, but the pointwise inverse to $\varphi$).
This proof shows us something slightly stronger: if $G$ is an arbitrary group, and $N\lhd G$ arbitrary, then the union of path components of all elements of $N$ is again a normal subgroup. In this case we have $N$ as the trivial subgroup, which is of course always normal.
