Permutation matrix properties proof

The problem asks to establish the following properties of $\displaystyle n \times n$ permutation matrices, for all $\displaystyle \sigma, \tau \in \mathcal{S}_n$.

1. $\displaystyle P_\sigma = I$ iff $\displaystyle \sigma(k) = k$ for all $\displaystyle k$, where $\displaystyle I$ denotes the $\displaystyle n \times n$ identity matrix.

2. $\displaystyle P_\sigma P_\tau = P_{\sigma \circ \tau}$.

3. $\displaystyle (P_\sigma)^{-1} = P_{\sigma^{-1}}$.

4. $\displaystyle P_\sigma$ is an orthogonal matrix, that is, $\displaystyle (P_\sigma)^{-1} = (P_\sigma)^T$.

Can someone pls check if my attempts below are correct proofs? Does the first one below qualify as a proof?

1. If $\displaystyle \sigma(k) = k$ for all $\displaystyle k$ then $\displaystyle P_\sigma A$ does not swap any rows of $\displaystyle A$ and therefore $\displaystyle P_\sigma A = A = I A$ which implies $\displaystyle P_\sigma = I$. If $\displaystyle P_\sigma = I$ then $\displaystyle P_\sigma A = I A = A$ meaning $\displaystyle \sigma(k) = k$ for all $\displaystyle k$.

2. $\displaystyle P_\sigma P_\tau A$ swaps $\displaystyle k$th row of $\displaystyle A$ with $\displaystyle \tau(k)$th row which is then swapped with $\displaystyle \sigma(\tau(k))$th row. So $\displaystyle k$th row is swapped with $\displaystyle (\sigma \circ \tau)(k)$th row which is what $\displaystyle P_{\sigma \circ \tau} A$ does. Therefore $\displaystyle P_\sigma P_\tau = P_{\sigma \circ \tau}$.

3. Property 2 says that $\displaystyle P_{\sigma}P_{\sigma^{-1}} = P_{\sigma \circ \sigma^{-1}}$. Because $\displaystyle (\sigma \circ \sigma^{-1})(k) = k$ for all $\displaystyle k$, property 1 says $\displaystyle P_{\sigma \circ \sigma^{-1}} = I$. Hence $\displaystyle P_{\sigma}P_{\sigma^{-1}} = I = P_{\sigma}(P_{\sigma})^{-1}$ and $\displaystyle (P_\sigma)^{-1} = P_{\sigma^{-1}}$.

4. If $\displaystyle [P_{\sigma}]_{ij} = 1$, then the $\displaystyle j$th row of $\displaystyle A$ will be the $\displaystyle i$th row of $\displaystyle P_{\sigma} A$. The inverse $\displaystyle P_{\sigma}^{-1}$ must copy the $\displaystyle i$th row back to the $\displaystyle j$th row and therefore $\displaystyle [P_{\sigma}^{-1}]_{ji} = [P_{\sigma}]_{ij}$. Let all other entries of $\displaystyle (P_{\sigma})^{-1}$ be zero, since all other $\displaystyle P_{\sigma}$ entries are also zero, we have $\displaystyle [P_{\sigma}^{-1}]_{ij} = [P_{\sigma}]_{ji} = [P_{\sigma}^T]_{ij}$.