Reflexive Generalized Inverse - Mathematics Stack Exchange
Sep 26, 2022 · 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 …
Prove that $o (a)=o (gag^ {-1})$ - Mathematics Stack Exchange
Sep 20, 2015 · Your proof of the second part works perfectly, moreover, you can simply omit the reasoning $ (gag^ {-1})^2=\cdots=e$ since this is exactly what you've done in part 1.
Explicitly write down $g\\in GL(n,\\mathbb{C})$ so that $gAg^{-1}$ is ...
One method is to explicitly write down gAg−1 g A g 1 and set the function in 2nd row, 1st column equal to zero (which is −a12g221 +g22(a11g21 −a22g21 +a21g22) = 0 a 12 g 21 2 + g 22 (a 11 g 21 a 22 …
abstract algebra - Centralizer and Normalizer as Group Action ...
Jan 3, 2019 · The stabilizer subgroup we defined above for this action on some set $A\subseteq G$ is the set of all $g\in G$ such that $gAg^ {-1} = A$ — which is exactly the normalizer subgroup $N_G (A)$!
Difference between a group normalizer and centralizer
Sep 27, 2015 · Let H is a Subgroup of G. Now if H is not normal if any element $ {g \in G}$ doesn't commute with H. Now we want to find if not all $ {g \in G}$, then which are the elements of G that …
Difference between left and right conjugation and commutativity
Jan 28, 2024 · In group theory, through the book of Dummitt and Foote, how is the idea of right and left conjugation different from commutation? The difference between them is not clear to me, as …
Proving $H\subset gHg^ {-1}$ without the normality condition
Dec 7, 2023 · No, but before I provide a counterexample, note that the map $\gamma_g=a\mapsto gag^ {-1}$ is a bijection at least, since it has an inverse in $\gamma_ {g^ {-1}}=a\mapsto g^ {-1}ag$.
Proving that $gHg^ {-1}$ is a subgroup of $G$
Jul 9, 2015 · $1) $$ (gag^ {-1})^ {-1}=g^ {-1^ {-1}}a^ {-1}g^ {-1}=ga^ {-1}g^ {-1}$ $2)$ $ ga (g^ {-1}g)bg^ {-1}=g (ab)g^ {-1}$ I'm stuck at this point, Is it correct so far? is ...
Let $G$ a group and $g,h∈G$. Prove that $|gh|=|hg|$.
(Two elements $a$ and $b$ are conjugate if there exists $g\in G$ such that $gag^ {-1}=b$.)
Conjugacy Classes of the Quaternion Group $Q$
Disclaimer: This is not exactly an explanation, but a relevant attempt at understanding conjugates and conjugate classes.