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)
Difference between left and right conjugation and commutativity 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 conjugation ap
Difference between a group normalizer and centralizer 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 commute with every element of H? they are normalizer of H i e , the elements of G that vote 'yes' for H when asked to commute Hence, $ {N_G (H)=\ {g \in G: gH=Hg }\}$ | Now Centralizer of an element $ {a \in G
Proving $H\subset gHg^ {-1}$ without the normality condition 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 $\gamma_ {g^ {-1}}=a\mapsto g^ {-1}ag$
Map between $SU (2)$ and $SO (3)$ - Mathematics Stack Exchange With all due respect to Manton, if that is indeed what he does, defining the map from $\mathrm {SU}_2$ to $\mathrm {SO}_3$ by this formula is a god-awful way to describe the map! Any element of $\mathrm {U}_2$ is a matrix of the form $$ A=\left (\begin {array} {cc} z_1 -\lambda\bar {w_1}\\ w_1 \lambda\bar {z_1} \end {array}\right) $$ where $|z_1|^2+|w_1|^2=1=|\lambda|^2$ Indeed the
abstract algebra - Show that conjugate by $g$ is isomorphism . . . All is fine An alternative way to establish bijectivity might be the observation that $\sigma_g\circ\sigma_h=\sigma _ {gh}$ (a useful fact on its own!) and therefore $\sigma_ {g^ {-1}}\circ \sigma_ {g}=\sigma_ {g}\circ \sigma_ {g^ {-1}}=\operatorname {id}_G$ - And a map with left and right inverse map is bijective Then again, this does not reall ydiffer from what you wrote, does it?