Math Problem Statement

Help me find number 2

Help me find number 2

Solution

Let's solve problem number 2 step by step.

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

Solution:

  • To prove that ρa\rho_a is a permutation, we need to show that ρa\rho_a is bijective (both injective and surjective):
    1. Injective:
      Suppose ρa(x1)=ρa(x2)\rho_a(x_1) = \rho_a(x_2).
      This implies ax1=ax2ax_1 = ax_2. Since aGa \in G and GG is a group, aa has an inverse a1a^{-1}. Multiplying both sides of the equation by a1a^{-1}, we get:
      x1=x2.x_1 = x_2.
      Therefore, ρa\rho_a is injective.
    2. Surjective:
      Let yGy \in G. We need to show that there exists xGx \in G such that ρa(x)=y\rho_a(x) = y, i.e., ax=yax = y.
      Since aa is in GG and GG is a group, the inverse a1a^{-1} exists. Multiply both sides of ax=yax = y by a1a^{-1}:
      x=a1y.x = a^{-1}y.
      Since a1yGa^{-1}y \in G, this shows ρa\rho_a is surjective.

Hence, ρa\rho_a is bijective and therefore a permutation on GG.


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

Solution:

  • SGS_G is the group of all permutations on GG.
  • To prove GG'' is a subgroup of SGS_G, we use the subgroup criterion:
    1. Closure under composition:
      Let ρa,ρbG\rho_a, \rho_b \in G''. Then for any xGx \in G,
      (ρaρb)(x)=ρa(ρb(x))=ρa(bx)=a(bx)=(ab)x.(\rho_a \circ \rho_b)(x) = \rho_a(\rho_b(x)) = \rho_a(bx) = a(bx) = (ab)x.
      Since abGab \in G, ρabG\rho_{ab} \in G''. Thus, GG'' is closed under composition.
    2. Identity element:
      The identity map in SGS_G is ρe(x)=ex=x\rho_e(x) = ex = x, where ee is the identity element of GG. Since eGe \in G, ρeG\rho_e \in G''.
    3. Inverses:
      For ρaG\rho_a \in G'', we need to show that ρa1G\rho_a^{-1} \in G''. Note that for any xGx \in G,
      ρa1(x)=a1x.\rho_a^{-1}(x) = a^{-1}x.
      Since a1Ga^{-1} \in G, ρa1G\rho_{a^{-1}} \in G''.

Therefore, GG'' is a subgroup of SGS_G.


Problem 2(c): Show that the map μ:GG\mu: G \to G'' defined by μ(a)=ρa1,aG\mu(a) = \rho_a^{-1}, \forall a \in G, is an isomorphism.

Solution:

  • To prove μ\mu is an isomorphism, we need to show that μ\mu is a homomorphism, injective, and surjective:
    1. Homomorphism:
      For a,bGa, b \in G,
      μ(ab)=ρab1.\mu(ab) = \rho_{ab}^{-1}.
      By definition of ρa\rho_a,
      ρab(x)=(ab)x,\rho_{ab}(x) = (ab)x,
      so the inverse is ρab1(x)=(ab)1x=b1a1x\rho_{ab}^{-1}(x) = (ab)^{-1}x = b^{-1}a^{-1}x. On the other hand,
      μ(a)μ(b)=ρa1ρb1,\mu(a) \circ \mu(b) = \rho_a^{-1} \circ \rho_b^{-1},
      and for any xGx \in G,
      (ρa1ρb1)(x)=ρa1(ρb1(x))=ρa1(b1x)=a1(b1x)=(ab)1x.(\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, μ(ab)=μ(a)μ(b)\mu(ab) = \mu(a) \circ \mu(b), so μ\mu is a homomorphism.
    2. Injective:
      Suppose μ(a)=μ(b)\mu(a) = \mu(b). This implies ρa1=ρb1\rho_a^{-1} = \rho_b^{-1}, which means for all xGx \in G,
      [

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Math Problem Analysis

Mathematical Concepts

Group Theory
Permutations
Subgroups
Isomorphisms

Formulas

ρ_a(x) = ax for all x ∈ G
Composition of functions: (ρ_a ∘ ρ_b)(x) = ρ_a(ρ_b(x))
Identity map: ρ_e(x) = x
Inverse map: ρ_a^{-1}(x) = a^{-1}x

Theorems

Cayley's Theorem
Group Isomorphism Properties
Subgroup Closure Criterion

Suitable Grade Level

Undergraduate (Advanced Mathematics)