On path-components of a subset of the set of all real unipotent matrices 
What are the path-components of the following set?
$$P = \{ A \in M_n(\mathbb R) \mid A^k=I_n\}$$

My approach:
Here it's seen that $P$ is not connected as $P_1 = \operatorname {det}^{-1}\{1\}$ and $P_2 =\operatorname {det}^{-1}\{-1\}$ is a seperation of $P$ ; but I think these above two sets are path-connected ; and I think we have to use the Jordan canonical form of those matrices $A .$
But, I am unable to think further .
Also, the matrces $\{A \in M_n(\mathbb C) | A^k =I_n\}$ are diagonalizable over $\mathbb C$, but what about in $\mathbb R ??$
 A: For even $n$, the my approach hinges on this question of mine. For odd $n$ the solution is as follows.
Note that the possible Jordan canonical forms (over $\mathbb C$) are just the diagonal matrices with $k$-th roots of unity on their diagonals.
Let $\mathcal P = \{A \in M_n(\mathbb R) \mid A^k = I\}.$ Then we have that $A$ and $B$ lie in the same path-component of $\mathcal P$ if and only if they have the same Jordan canonical form.
Proof.
First, note that all paths in the set of $k$-th roots of unity (in $\mathbb C$) are constant. Since the dependency of a matrix to its eigenvalues is continuous, we have that a path in the set of unipotent matrices cannot include matrices of distinct Jordan canonical forms. 
Conversely, suppose $A$ and $B$ have the same Jordan canonical form. Then there exists a real matrix $P$ such that $B = PAP^{-1}$ (this is because whenever real matrices are conjugate over $\mathbb C$ then they are also conjugate over $\mathbb R$). 
Suppose $\det P < 0$. Then there exists a real matrix $P'$ (namely $P' = -P$) such that $\det P' > 0$ and $B = P'AP'^{-1}$. 
We can thus assume that $\det P > 0$.
Then we have that $P = e^Q$ for some real matrix $Q$ (since the image of the exponential map is the path-component of the identity in $\text{GL}_n(\mathbb R)$). Then the path $$t \mapsto e^{tQ}Ae^{-tQ}$$
is a path connecting $A$ to $PAP^{-1} = B$ inside $\mathcal P$.
