1. **Problem Statement:** We consider the group of isometries $\mathrm{Isom}(\mathbb{R}^n)$ under composition. Define: - $H = \{\varphi \in \mathrm{Isom}(\mathbb{R}^n) : \varphi(0) = 0\}$. - For $d \in \mathbb{R}^n$, $T_d : \mathbb{R}^n \to \mathbb{R}^n$ by $T_d(x) = x + d$. - $K = \{T_d : d \in \mathbb{R}^n\} \subseteq \mathrm{Isom}(\mathbb{R}^n)$. We need to prove: (a) $H$ and $K$ are subgroups of $\mathrm{Isom}(\mathbb{R}^n)$. (b) For $\varphi \in \mathrm{Isom}(\mathbb{R}^n)$ and $T_d \in K$, $\varphi \circ T_d \circ \varphi^{-1} \in K$ and conclude $K \triangleleft \mathrm{Isom}(\mathbb{R}^n)$. (c) $H \cap K$ is trivial and $HK = \mathrm{Isom}(\mathbb{R}^n)$. (d) Explain why $\mathrm{Isom}(\mathbb{R}^n) \cong K \rtimes_\theta H$ for a homomorphism $\theta : H \to \mathrm{Aut}(K)$. 2. **(a) Prove $H$ and $K$ are subgroups:** - To show $H$ is a subgroup, check closure, identity, and inverses: - Closure: If $\varphi, \psi \in H$, then $(\varphi \circ \psi)(0) = \varphi(\psi(0)) = \varphi(0) = 0$, so $\varphi \circ \psi \in H$. - Identity: The identity map $\mathrm{id}$ satisfies $\mathrm{id}(0) = 0$, so $\mathrm{id} \in H$. - Inverses: If $\varphi \in H$, then $\varphi(0) = 0$. Since $\varphi$ is an isometry, it is bijective, so $\varphi^{-1}(0) = 0$, hence $\varphi^{-1} \in H$. - To show $K$ is a subgroup: - Closure: For $T_d, T_e \in K$, $T_d \circ T_e(x) = T_d(x + e) = x + e + d = T_{d+e}(x)$, so $T_d \circ T_e = T_{d+e} \in K$. - Identity: $T_0(x) = x + 0 = x$ is the identity map, so $T_0 \in K$. - Inverses: $T_d^{-1} = T_{-d}$ since $T_d \circ T_{-d} = T_0$. Thus, $H$ and $K$ are subgroups. 3. **(b) Conjugation of $K$ by $\varphi$ lies in $K$ and normality:** - Let $\varphi \in \mathrm{Isom}(\mathbb{R}^n)$ and $T_d \in K$. - Compute $\varphi \circ T_d \circ \varphi^{-1}(x) = \varphi(\varphi^{-1}(x) + d) = x + \varphi(d) - \varphi(0)$ since isometries are affine maps. - Since $\varphi$ is an isometry, it can be written as $\varphi(x) = A x + b$ with $A$ orthogonal and $b = \varphi(0)$. - Then $\varphi \circ T_d \circ \varphi^{-1}(x) = x + A d$. - Hence $\varphi \circ T_d \circ \varphi^{-1} = T_{A d} \in K$. - Since conjugation by any element of $\mathrm{Isom}(\mathbb{R}^n)$ sends $K$ to itself, $K$ is a normal subgroup: $K \triangleleft \mathrm{Isom}(\mathbb{R}^n)$. 4. **(c) Intersection and product:** - $H \cap K$ consists of isometries fixing 0 and translations. - A translation $T_d$ fixes 0 iff $T_d(0) = 0 + d = 0 \implies d = 0$. - So $H \cap K = \{T_0\} = \{\mathrm{id}\}$, the trivial subgroup. - To show $HK = \mathrm{Isom}(\mathbb{R}^n)$: - Any isometry $\varphi$ can be written as $\varphi(x) = A x + b$ with $A$ orthogonal and $b \in \mathbb{R}^n$. - Define $\psi(x) = A x$ which fixes 0, so $\psi \in H$. - Define $T_b(x) = x + b \in K$. - Then $\varphi = T_b \circ \psi \in HK$. - Hence $HK = \mathrm{Isom}(\mathbb{R}^n)$. 5. **(d) Semidirect product structure:** - Define $\theta : H \to \mathrm{Aut}(K)$ by $\theta(\varphi)(T_d) = \varphi \circ T_d \circ \varphi^{-1}$. - From (b), $\theta(\varphi)(T_d) = T_{A d}$ where $\varphi(x) = A x + b$. - This is a group homomorphism from $H$ to automorphisms of $K$. - Since $H \cap K$ is trivial and $HK = \mathrm{Isom}(\mathbb{R}^n)$, the group $\mathrm{Isom}(\mathbb{R}^n)$ is isomorphic to the semidirect product $K \rtimes_\theta H$. **Final conclusion:** $$\mathrm{Isom}(\mathbb{R}^n) \cong K \rtimes_\theta H,$$ where $K$ is the normal subgroup of translations and $H$ is the subgroup of isometries fixing the origin.