1. **Problem Statement:** Prove that every group $G$ is isomorphic to a group of permutations on $G$ itself, and that if $\varphi: G \to G'$ is an isomorphism, then its inverse $\varphi^{-1}$ is also an isomorphism from $G'$ onto $G$. 2. **Key Definitions and Formulas:** - A group $G$ is a set with an operation satisfying closure, associativity, identity, and inverse. - A permutation on $G$ is a bijection from $G$ to $G$. - An isomorphism $\varphi: G \to G'$ is a bijective homomorphism: $\varphi(ab) = \varphi(a)\varphi(b)$ for all $a,b \in G$. 3. **Step 1: Define the function $f_x$ for each $x \in G$:** $$f_x: G \to G, \quad f_x(a) = ax$$ 4. **Show $f_x$ is a permutation:** - *One-to-one:* If $f_x(a) = f_x(b)$, then $ax = bx \implies a = b$ by cancellation. - *Onto:* For any $y \in G$, choose $a = yx^{-1}$, then $f_x(a) = yx^{-1}x = y$. Thus, $f_x$ is bijection, i.e., a permutation. 5. **Define $G^x = \{ f_x : x \in G \}$ and show it forms a group under composition:** - *Closure:* $(f_a \circ f_b)(x) = f_a(f_b(x)) = f_a(bx) = a(bx) = (ab)x = f_{ab}(x)$, so $f_a \circ f_b = f_{ab} \in G^x$. - *Associativity:* Composition of functions is associative. - *Identity:* $f_e$ where $e$ is identity in $G$ acts as identity function. - *Inverse:* $f_a^{-1} = f_{a^{-1}}$ since $f_a \circ f_{a^{-1}} = f_e$. 6. **Define $\varphi: G \to G^x$ by $\varphi(a) = f_a$ and prove it is an isomorphism:** - *One-to-one:* $\varphi(a) = \varphi(b) \implies f_a = f_b \implies a = b$. - *Onto:* Every $f_a \in G^x$ has a preimage $a \in G$. - *Homomorphism:* $\varphi(ab) = f_{ab} = f_a \circ f_b = \varphi(a) \varphi(b)$. 7. **Conclusion:** $G$ is isomorphic to $G^x$, a group of permutations on $G$. 8. **Prove $\varphi^{-1}$ is an isomorphism from $G'$ onto $G$:** - Since $\varphi$ is bijection, $\varphi^{-1}$ exists. - For $x', y' \in G'$, write $x' = \varphi(x)$, $y' = \varphi(y)$ for some $x,y \in G$. - Then, $$\varphi^{-1}(x' y') = \varphi^{-1}(\varphi(x) \varphi(y)) = \varphi^{-1}(\varphi(xy)) = xy = \varphi^{-1}(x') \varphi^{-1}(y')$$ - So $\varphi^{-1}$ preserves the group operation. - $\varphi^{-1}$ is bijection by definition. 9. **Final statement:** The inverse of an isomorphism is also an isomorphism. **Answer:** Every group $G$ is isomorphic to a group of permutations on $G$, and the inverse of an isomorphism $\varphi$ is also an isomorphism.