Let's solve problem number 2 step by step.

Problem 2(a): Show that $\rho_a: G \to G$ defined by $\rho_a(x) = ax$, $\forall x \in G$, is a permutation on $G$.

Solution:

• To prove that $\rho_a$ is a permutation, we need to show that $\rho_a$ is bijective (both injective and surjective):
  1. Injective:
     Suppose $\rho_a(x_1) = \rho_a(x_2)$.
     This implies $ax_1 = ax_2$. Since $a \in G$ and $G$ is a group, $a$ has an inverse $a^{-1}$. Multiplying both sides of the equation by $a^{-1}$, we get:
     $x_1 = x_2.$
     Therefore, $\rho_a$ is injective.
  2. Surjective:
     Let $y \in G$. We need to show that there exists $x \in G$ such that $\rho_a(x) = y$, i.e., $ax = y$.
     Since $a$ is in $G$ and $G$ is a group, the inverse $a^{-1}$ exists. Multiply both sides of $ax = y$ by $a^{-1}$:
     $x = a^{-1}y.$
     Since $a^{-1}y \in G$, this shows $\rho_a$ is surjective.

Hence, $\rho_a$ is bijective and therefore a permutation on $G$.

Problem 2(b): Show that $G'' = \{\rho_a \mid a \in G\}$ is a subgroup of $S_G$.

Solution:

• $S_G$ is the group of all permutations on $G$.
• To prove $G''$ is a subgroup of $S_G$, we use the subgroup criterion:
  1. Closure under composition:
     Let $\rho_a, \rho_b \in G''$. Then for any $x \in G$,
     $(\rho_a \circ \rho_b)(x) = \rho_a(\rho_b(x)) = \rho_a(bx) = a(bx) = (ab)x.$
     Since $ab \in G$, $\rho_{ab} \in G''$. Thus, $G''$ is closed under composition.
  2. Identity element:
     The identity map in $S_G$ is $\rho_e(x) = ex = x$, where $e$ is the identity element of $G$. Since $e \in G$, $\rho_e \in G''$.
  3. Inverses:
     For $\rho_a \in G''$, we need to show that $\rho_a^{-1} \in G''$. Note that for any $x \in G$,
     $\rho_a^{-1}(x) = a^{-1}x.$
     Since $a^{-1} \in G$, $\rho_{a^{-1}} \in G''$.

Therefore, $G''$ is a subgroup of $S_G$.

Problem 2(c): Show that the map $\mu: G \to G''$ defined by $\mu(a) = \rho_a^{-1}, \forall a \in G$, is an isomorphism.

Solution:

• To prove $\mu$ is an isomorphism, we need to show that $\mu$ is a homomorphism, injective, and surjective:
  1. Homomorphism:
     For $a, b \in G$,
     $\mu(ab) = \rho_{ab}^{-1}.$
     By definition of $\rho_a$,
     $\rho_{ab}(x) = (ab)x,$
     so the inverse is $\rho_{ab}^{-1}(x) = (ab)^{-1}x = b^{-1}a^{-1}x$. On the other hand,
     $\mu(a) \circ \mu(b) = \rho_a^{-1} \circ \rho_b^{-1},$
     and for any $x \in G$,
     $(\rho_a^{-1} \circ \rho_b^{-1})(x) = \rho_a^{-1}(\rho_b^{-1}(x)) = \rho_a^{-1}(b^{-1}x) = a^{-1}(b^{-1}x) = (ab)^{-1}x.$
     Hence, $\mu(ab) = \mu(a) \circ \mu(b)$, so $\mu$ is a homomorphism.
  2. Injective:
     Suppose $\mu(a) = \mu(b)$. This implies $\rho_a^{-1} = \rho_b^{-1}$, which means for all $x \in G$,
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