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• φ maps S onto S' in a one-to-one fashion, meaning that each element u in S' has a unique corresponding element a in S such that φ(a) = u. Since a is unique, the inverse map φ-1(u) = a from S' to S is well defined and unique. This shows φ-1 is …

  in Abstract Algebra Isomorphic? Comment by rodolfo-riverol June 2021
• φ maps S onto S' in a one-to-one fashion, meaning that each element u in S' has a unique corresponding element a in S such that φ(a) = u. Since a is unique, the inverse map φ-1(u) = a from S' to S is well defined and unique. This shows φ-1 is …

  in Abstract Algebra Isomorphic? Comment by rodolfo-riverol June 2021
• Determine the units of Z[ (-1+root(-3))/2 ]. Let a + b(-1+root(-3))/2 be a unit. Taking norms a^2 -ab + b^2 =1, so that 4a^2 - 4ab + 4b^2 = 4, and finally (2a-b)^2 = 4-3b^2. The possible values for b are 0, +/-1. If b = 0 then a = +/-1, if b = 1 …

  in Quadratic Integers & Associates? Comment by rodolfo-riverol June 2021
• If N(a)x = N(a)y then N(a) = N(a)yx^-1 which means that yx^-1 is in N(a) so a = (yx^-1)^-1a(yx^-1) = xy^-1ayx^-1 ==> a = xy^-1ayx^-1 ==> x^-1ax = y^-1ay. Working backwards you get the reverse. The conjugacy class of a, Cl(a) = {gag^-1 for …

  in Abstract Linear Algebra Problem? Comment by rodolfo-riverol May 2021
• If N(a)x = N(a)y then N(a) = N(a)yx^-1 which means that yx^-1 is in N(a) so a = (yx^-1)^-1a(yx^-1) = xy^-1ayx^-1 ==> a = xy^-1ayx^-1 ==> x^-1ax = y^-1ay. Working backwards you get the reverse. The conjugacy class of a, Cl(a) = {gag^-1 for …

  in Abstract Linear Algebra Problem? Comment by rodolfo-riverol May 2021
• If N(a)x = N(a)y then N(a) = N(a)yx^-1 which means that yx^-1 is in N(a) so a = (yx^-1)^-1a(yx^-1) = xy^-1ayx^-1 ==> a = xy^-1ayx^-1 ==> x^-1ax = y^-1ay. Working backwards you get the reverse. The conjugacy class of a, Cl(a) = {gag^-1 for …

  in Abstract Linear Algebra Problem? Comment by rodolfo-riverol May 2021
• If N(a)x = N(a)y then N(a) = N(a)yx^-1 which means that yx^-1 is in N(a) so a = (yx^-1)^-1a(yx^-1) = xy^-1ayx^-1 ==> a = xy^-1ayx^-1 ==> x^-1ax = y^-1ay. Working backwards you get the reverse. The conjugacy class of a, Cl(a) = {gag^-1 for …

  in Abstract Linear Algebra Problem? Comment by rodolfo-riverol May 2021
• Note that [k(a,b):k] = [k(a,b):k(a)][k(a):k] = [k(a,b):k(a)]m, which shows that m divides [k(a,b):k]. Symmetrically [k(a,b):k] = [k(a,b):k(b)][k(b):k] = [k(a,b):k(b)]n, showing that n divides [k(a,b):k]. Since m and n are relatively prime then mn …

  in Math (abstract algebra)? Comment by rodolfo-riverol May 2021
• Note that [k(a,b):k] = [k(a,b):k(a)][k(a):k] = [k(a,b):k(a)]m, which shows that m divides [k(a,b):k]. Symmetrically [k(a,b):k] = [k(a,b):k(b)][k(b):k] = [k(a,b):k(b)]n, showing that n divides [k(a,b):k]. Since m and n are relatively prime then mn …

  in Math (abstract algebra)? Comment by rodolfo-riverol May 2021
• Note that [k(a,b):k] = [k(a,b):k(a)][k(a):k] = [k(a,b):k(a)]m, which shows that m divides [k(a,b):k]. Symmetrically [k(a,b):k] = [k(a,b):k(b)][k(b):k] = [k(a,b):k(b)]n, showing that n divides [k(a,b):k]. Since m and n are relatively prime then mn …

  in Math (abstract algebra)? Comment by rodolfo-riverol May 2021
• If N(a)x = N(a)y then N(a) = N(a)yx^-1 which means that yx^-1 is in N(a) so a = (yx^-1)^-1a(yx^-1) = xy^-1ayx^-1 ==> a = xy^-1ayx^-1 ==> x^-1ax = y^-1ay. Working backwards you get the reverse. The conjugacy class of a, Cl(a) = {gag^-1 for …

  in Abstract Linear Algebra Problem? Comment by rodolfo-riverol May 2021
• If N(a)x = N(a)y then N(a) = N(a)yx^-1 which means that yx^-1 is in N(a) so a = (yx^-1)^-1a(yx^-1) = xy^-1ayx^-1 ==> a = xy^-1ayx^-1 ==> x^-1ax = y^-1ay. Working backwards you get the reverse. The conjugacy class of a, Cl(a) = {gag^-1 for …

  in Abstract Linear Algebra Problem? Comment by rodolfo-riverol May 2021
• Determine the units of Z[ (-1+root(-3))/2 ]. Let a + b(-1+root(-3))/2 be a unit. Taking norms a^2 -ab + b^2 =1, so that 4a^2 - 4ab + 4b^2 = 4, and finally (2a-b)^2 = 4-3b^2. The possible values for b are 0, +/-1. If b = 0 then a = +/-1, if b = 1 …

  in Quadratic Integers & Associates? Comment by rodolfo-riverol May 2021