[proofplan]
We use the defining property $Q^\top Q = I$ of an orthogonal matrix throughout. Part 1 follows directly from this definition. Part 2 is proved by inserting $Q^\top Q = I$ into the inner product. Part 3 is the special case $w = v$ of Part 2. Part 4 follows from the multiplicativity of the determinant applied to $Q^\top Q = I$. Part 5 verifies the group axioms by checking the defining property for products and inverses.
[/proofplan]
[step:Establish invertibility and $Q^{-1} = Q^\top$ from the definition]
Recall that $Q \in \mathbb{R}^{n \times n}$ is orthogonal means $Q^\top Q = I_n$, where $I_n$ is the $n \times n$ identity matrix. This says precisely that $Q^\top$ is a left inverse of $Q$. Since $Q$ is a square matrix and has a left inverse, $Q$ is invertible and $Q^{-1} = Q^\top$.
It follows that $QQ^\top = I_n$ as well (a two-sided inverse equals the one-sided inverse for square matrices).
[/step]
[step:Prove the inner product identity $\langle Qv, Qw \rangle = \langle v, w \rangle$]
For $v, w \in \mathbb{R}^n$, the Euclidean inner product satisfies $\langle x, y \rangle = x^\top y$ (viewing vectors as column vectors). Therefore
\begin{align*}
\langle Qv, Qw \rangle &= (Qv)^\top (Qw) = v^\top Q^\top Q\, w = v^\top I_n\, w = v^\top w = \langle v, w \rangle,
\end{align*}
where we used $(Qv)^\top = v^\top Q^\top$ and $Q^\top Q = I_n$.
[guided]
We want to show that $Q$ preserves the inner product. On $\mathbb{R}^n$, the Euclidean inner product is $\langle x, y \rangle = x^\top y = \sum_{i=1}^n x_i y_i$. We compute:
\begin{align*}
\langle Qv, Qw \rangle &= (Qv)^\top (Qw).
\end{align*}
The transpose of a product reverses the order: $(Qv)^\top = v^\top Q^\top$. Substituting:
\begin{align*}
\langle Qv, Qw \rangle &= v^\top Q^\top Q\, w.
\end{align*}
Now we use the defining property of orthogonality, $Q^\top Q = I_n$:
\begin{align*}
\langle Qv, Qw \rangle &= v^\top I_n\, w = v^\top w = \langle v, w \rangle.
\end{align*}
This shows that $Q$ is an isometry of the inner product. The argument is essentially a matrix reformulation of orthogonality: $Q^\top Q = I_n$ is exactly the statement that the columns of $Q$ form an orthonormal set, and inner-product preservation is the coordinate-free consequence.
[/guided]
[/step]
[step:Deduce the norm identity $\|Qv\| = \|v\|$ as a special case]
Setting $w = v$ in Part 2:
\begin{align*}
\|Qv\|^2 &= \langle Qv, Qv \rangle = \langle v, v \rangle = \|v\|^2.
\end{align*}
Since norms are non-negative, taking square roots gives $\|Qv\| = \|v\|$ for all $v \in \mathbb{R}^n$.
[/step]
[step:Compute $|\det Q| = 1$ from $Q^\top Q = I_n$]
Applying the determinant to both sides of $Q^\top Q = I_n$ and using the multiplicativity of the determinant:
\begin{align*}
\det(Q^\top Q) &= \det(I_n) = 1.
\end{align*}
Since $\det(Q^\top) = \det(Q)$ (the determinant of a matrix equals the determinant of its transpose) and $\det(Q^\top Q) = \det(Q^\top) \det(Q)$:
\begin{align*}
(\det Q)^2 &= 1.
\end{align*}
Therefore $\det Q = \pm 1$, i.e., $|\det Q| = 1$.
[/step]
[step:Verify that $O(n)$ is a group under matrix multiplication]
We check the four group axioms for $O(n) := \{Q \in \mathbb{R}^{n \times n} : Q^\top Q = I_n\}$.
**Closure.** Let $Q_1, Q_2 \in O(n)$. Then
\begin{align*}
(Q_1 Q_2)^\top (Q_1 Q_2) &= Q_2^\top Q_1^\top Q_1 Q_2 = Q_2^\top I_n Q_2 = Q_2^\top Q_2 = I_n,
\end{align*}
so $Q_1 Q_2 \in O(n)$.
**Identity.** $I_n^\top I_n = I_n$, so $I_n \in O(n)$.
**Inverses.** For $Q \in O(n)$, Part 1 gives $Q^{-1} = Q^\top$. We verify $Q^\top \in O(n)$: $(Q^\top)^\top Q^\top = Q Q^\top = I_n$ (using the identity $QQ^\top = I_n$ established in Part 1). So $Q^{-1} = Q^\top \in O(n)$.
**Associativity.** Matrix multiplication is associative.
Therefore $O(n)$ is a group under multiplication.
[guided]
We must verify the group axioms for $O(n)$, the set of all $n \times n$ orthogonal matrices.
**Closure.** Let $Q_1, Q_2 \in O(n)$, meaning $Q_1^\top Q_1 = I_n$ and $Q_2^\top Q_2 = I_n$. Is $Q_1 Q_2$ orthogonal? We check:
\begin{align*}
(Q_1 Q_2)^\top (Q_1 Q_2) &= Q_2^\top Q_1^\top Q_1 Q_2,
\end{align*}
where we used the transpose reversal rule $(AB)^\top = B^\top A^\top$. Now substitute $Q_1^\top Q_1 = I_n$:
\begin{align*}
Q_2^\top Q_1^\top Q_1 Q_2 &= Q_2^\top I_n Q_2 = Q_2^\top Q_2 = I_n.
\end{align*}
So $Q_1 Q_2 \in O(n)$.
**Identity.** The identity matrix $I_n$ satisfies $I_n^\top I_n = I_n$, so $I_n \in O(n)$ and acts as the group identity.
**Inverses.** For $Q \in O(n)$, Part 1 established that $Q^{-1} = Q^\top$ and that $QQ^\top = I_n$. We need $Q^{-1} = Q^\top$ to itself be in $O(n)$, i.e., $(Q^\top)^\top Q^\top = I_n$. Since $(Q^\top)^\top = Q$, this reduces to $QQ^\top = I_n$, which holds. So $Q^{-1} \in O(n)$.
**Associativity.** Matrix multiplication is associative (this is a standard property of matrix algebra, not specific to $O(n)$).
All four axioms are satisfied, so $O(n)$ is a group under multiplication. Note that $O(n)$ is a subgroup of $GL(n, \mathbb{R})$, the general linear group of invertible $n \times n$ real matrices.
[/guided]
[/step]
