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- Prove that $o(a)=o(gag^{-1})$ - Mathematics Stack Exchange
Your proof of the second part works perfectly, moreover, you can simply omit the reasoning (gag−1)2 = ⋯ = e (g a g 1) 2 = = e since this is exactly what you've done in part 1
- abstract algebra - Proving that $gHg^ {-1}$ is a subgroup of $G . . .
1) (gag−1)−1 ∈gHg−1 2) (gag−1) ⋅ (gbg−1) ∈gHg−1 1) (g a g − 1) − 1 ∈ g H g − 1 2) (g a g − 1) ⋅ (g b g − 1) ∈ g H g − 1 Since that is what your attempt proves Now having this you only need to check that e ∈ gHg−1 e ∈ g H g − 1 to conclude that indeed you have a subgroup ! – Alonso Delfín Commented
- Let $a \\in G$. Show that for any $g \\in G$, $gC(a)g^{-1} = C(gag^{-1})$.
Try checking if the element ghg−1 g h g 1 you thought of is in C(gag−1) C (g a g 1) and then vice versa
- 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)
- $G$ 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
- 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 σg: G → G σ
- Centralizer : center subgroup - Mathematics Stack Exchange
The normalizer works "the other way around" to what you may think it does The normalizer of A A in G G is the largest subgroup of G G that contains A A and in which A A is normal That is, NG(A) = {g ∈ G: gA = Ag} N G (A) = {g ∈ G: g A = A g} By its very definition, A A is normal in NG(A) N G (A) However, NG(A) N G (A) may not be a normal subgroup of G G, it seems you already have
- Difference between a group normalizer and centralizer
14 I would say (less precisely, but correctly) like this: g g is in NG(A) N G (A) means gag−1 = g a g − 1 = some a′ a ′ in A (a ∈ A a ∈ A) g g is in CG(A) C G (A) means gag−1 = g a g − 1 = same a a in A (a ∈ A a ∈ A) We should note that although there is difference between these two notions, there is also a relation
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