1. **Problem statement:** We are given groups $A$ and $B$, a homomorphism $\theta : A \to \mathrm{Aut}(B)$, and a set $B \times_\theta A$ with operation defined by $$ (b,a)(b',a') := (b \theta_a(b'), aa'). $$ We need to show: (a) $B \times_\theta A$ is a group. (b) $B \times \{1\}$ is a normal subgroup of $B \times_\theta A$. (c) For subgroups $H,K$ of $G$ with $K \triangleleft G$, $HK=G$, and $H \cap K = \{1\}$, define $\theta : H \to \mathrm{Aut}(K)$ by $\theta_h(k) = hkh^{-1}$. Show $\theta$ is a homomorphism. (d) Prove $G \cong K \times_\theta H$. 2. **Step (a): Show $B \times_\theta A$ is a group.** - **Closure:** For $(b,a),(b',a') \in B \times A$, $(b,a)(b',a') = (b \theta_a(b'), aa')$ is in $B \times A$ since $b \theta_a(b') \in B$ and $aa' \in A$. - **Associativity:** For $(b,a),(b',a'),(b'',a'') \in B \times A$, check $$ ((b,a)(b',a'))(b'',a'') = (b \theta_a(b'), aa')(b'',a'') = (b \theta_a(b') \theta_{aa'}(b''), aa'a''). $$ On the other hand, $$ (b,a)((b',a')(b'',a'')) = (b,a)(b' \theta_{a'}(b''), a'a'') = (b \theta_a(b' \theta_{a'}(b'')), a a' a''). $$ Since $\theta_a$ is an automorphism, $$ \theta_a(b' \theta_{a'}(b'')) = \theta_a(b') \theta_a(\theta_{a'}(b'')) = \theta_a(b') \theta_{a a'}(b''). $$ Thus, $$ (b,a)((b',a')(b'',a'')) = (b \theta_a(b') \theta_{a a'}(b''), a a' a'') $$ which equals the previous expression, so associativity holds. - **Identity:** The identity element is $(1_B,1_A)$ since $$ (b,a)(1_B,1_A) = (b \theta_a(1_B), a 1_A) = (b 1_B, a) = (b,a), $$ $$ (1_B,1_A)(b,a) = (1_B \theta_{1_A}(b), 1_A a) = (1_B b, a) = (b,a). $$ - **Inverses:** For $(b,a)$, define inverse $$ (b,a)^{-1} = (\theta_{a^{-1}}(b^{-1}), a^{-1}). $$ Check $$ (b,a)(\theta_{a^{-1}}(b^{-1}), a^{-1}) = (b \theta_a(\theta_{a^{-1}}(b^{-1})), a a^{-1}) = (b b^{-1}, 1_A) = (1_B,1_A). $$ Similarly, $$ (\theta_{a^{-1}}(b^{-1}), a^{-1})(b,a) = (\theta_{a^{-1}}(b^{-1}) \theta_{a^{-1}}(b), a^{-1} a) = (1_B,1_A). $$ Thus, $B \times_\theta A$ is a group. 3. **Step (b): Show $B \times \{1\}$ is normal in $B \times_\theta A$.** - $B \times \{1\}$ is a subgroup since for $(b,1),(b',1)$, $$ (b,1)(b',1) = (b \theta_1(b'), 1) = (b b', 1), $$ and inverses are $(b,1)^{-1} = (b^{-1},1)$. - To show normality, for any $(b,a) \in B \times_\theta A$ and $(b',1) \in B \times \{1\}$, compute conjugation: $$ (b,a)(b',1)(b,a)^{-1} = (b,a)(b',1)(\theta_{a^{-1}}(b^{-1}), a^{-1}) = (b \theta_a(b'), a)(\theta_{a^{-1}}(b^{-1}), a^{-1}) = (b \theta_a(b') \theta_a(\theta_{a^{-1}}(b^{-1})), a a^{-1}) = (b \theta_a(b') b^{-1}, 1). $$ Since $b \theta_a(b') b^{-1} \in B$, the conjugate lies in $B \times \{1\}$, so it is normal. 4. **Step (c): Show $\theta : H \to \mathrm{Aut}(K)$ defined by $\theta_h(k) = h k h^{-1}$ is a homomorphism.** - For $h,h' \in H$, $k \in K$, $$ \theta_{h h'}(k) = (h h') k (h h')^{-1} = h (h' k h'^{-1}) h^{-1} = \theta_h(\theta_{h'}(k)). $$ Thus, $$ \theta_{h h'} = \theta_h \circ \theta_{h'}, $$ so $\theta$ is a group homomorphism. 5. **Step (d): Prove $G \cong K \times_\theta H$.** - Define map $$ \varphi : K \times_\theta H \to G, \quad (k,h) \mapsto k h. $$ - **Well-defined:** Since $K,H \leq G$, $k h \in G$. - **Homomorphism:** For $(k,h),(k',h') \in K \times_\theta H$, $$ \varphi((k,h)(k',h')) = \varphi(k \theta_h(k'), h h') = k (h k' h^{-1}) h h' = k h k' h^{-1} h h' = k h k' h'. $$ On the other hand, $$ \varphi(k,h) \varphi(k',h') = (k h)(k' h') = k h k' h'. $$ So $\varphi$ is a homomorphism. - **Surjective:** Since $G = H K$ and $H K = G$, every $g \in G$ can be written as $g = h k$ for some $h \in H$, $k \in K$. Because $H K = G$ and $H \cap K = \{1\}$, the decomposition is unique and $g = k h$ for some $k \in K$, $h \in H$ (rearranged). - **Injective:** Suppose $\varphi(k,h) = 1_G$, then $k h = 1_G$, so $h = k^{-1}$. Since $h \in H$ and $k^{-1} \in K$, and $H \cap K = \{1\}$, it follows $h = 1$, $k = 1$. So kernel is trivial. Thus, $\varphi$ is an isomorphism. **Final answers:** (a) $B \times_\theta A$ is a group. (b) $B \times \{1\}$ is a normal subgroup. (c) $\theta$ is a homomorphism. (d) $G \cong K \times_\theta H$.