Prove that $o(a)=o(gag^{-1})$ - Mathematics Stack Exchange Let G G be a group and a ∈ G a ∈ G Prove that o(a) = o(gag−1) o (a) = o (g a g − 1) for every element of order 2 2 in G G If a be the only element of order 2 2 in G G deduce that a commutes with every element of G G Approach: Let o(a) = n o (a) = n, then an = e a n = e Now
$G$ is finite, $A \\leq G$ and all double cosets $AxA$ have the same . . . Closed 8 years ago If G G is a finite group and A A is a subgroup of G G such that all double cosets AxA A x A have the same number of elements, show that gAg−1 = A g A g − 1 = A for all g ∈ G g ∈ G Here is my attempt, I guess it's correct but please verify it
abstract algebra - $x$ conjugate to $y$ in a group $G$ is an . . . I am writing this answer largely to push through that, Group theory is not a collection of discrete facts, but really a continuum of ideas about symmetry in Mathematics So, having solved this exercise, you can ask yourself the following questions What is the number of equivalence classes (or since the relation is conjugation, conjugacy classes)? Answer This may not have a nice answer for
Let $a$ be an element of a group. Show that $a$ and $a^{-1}$ have the . . . I think one way to solve it is by considering the conjugation of a a and a−1 a − 1 using g ∈ G g ∈ G That is, ag = gag−1 a g = g a g − 1 and (a−1)g = ga−1g−1 (a − 1) g = g a − 1 g − 1 Then, if n n is the order of a a we have (ag)n = e (a g) n = e; since, (ag)n = gag−1gag−1 ⋯g−1gag−1 = (a g) n = g a g − 1 g a g − 1 ⋯ g − 1 g a g − 1 = = gaegaea ⋯
abstract algebra - Centralizer and Normalizer as Group Action . . . The stabilizer subgroup we defined above for this action on some set A ⊆ G A ⊆ G is the set of all g ∈ G g ∈ G such that gAg−1 = A g A g − 1 = A — which is exactly the normalizer subgroup NG(A) N G (A)! Thus we know that the normalizer is a subgroup because stabilizers are
Reflexive Generalized Inverse - Mathematics Stack Exchange Definition: G is a generalized inverse of A if and only if AGA=A G is said to be reflexive if and only if GAG=G I was trying to solve the problem: If A is a matrix and G be it's generalized inverse then G is reflexive if and only if rank (A)=rank (G)
Let $G$ be a group, $a \\in G$. Prove that for all $g \\in G$, $|a . . . I am stuck on this problem Let's just consider the case where |a| | a | is finite So |a| = n | a | = n where n n is the least positive integer such that an = e a n = e First I need to show that |g−1ag| ≤ n | g − 1 a g | ≤ n I notice that g−1ang =g−1eg =g−1g = e g − 1 a n g = g − 1 e g = g − 1 g = e But does this imply that (g−1ag)n = e (g − 1 a g) n = e? If so, why
by $g$ is isomorphism - Mathematics Stack Exchange For fixed g ∈ G g ∈ G, prove that conjugation by g g is an isomorphism from G G onto itself (i e an automorphism of G G) Deduce that x x and gxg−1 g x g − 1 have that same order for all x ∈ G x ∈ G and that for any subset A A of G G, |A| = |gAg−1| | A | = | g A g − 1 | (here gAg−1 = {gag−1|a ∈ A} g A g − 1 = {g a g − 1 | a ∈ A} ) Take g ∈ G g ∈ G We show that