Prove that $o (a)=o (gag^ {-1})$ for every element of order $2$ in $G$. If a be the only element of order $2$ in $G$ deduce that a commutes with every element of $G$
Try checking if the element $ghg^ {-1}$ you thought of is in $C (gag^ {-1})$ and then vice versa.
Let $a \in G$. Show that for any $g \in G$, $gC (a)g^ {-1} = C (gag ...
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).
So if you find a single eigenvector $v$ for $A$, and let $g^ {-1}$ be any invertible $2$-by-$2$ matrix whose first column is $v$, then $gAg^ {-1}$ is upper triangular.
$1) $$ (gag^ {-1})^ {-1}=g^ {-1^ {-1}}a^ {-1}g^ {-1}=ga^ {-1}g^ {-1}$ $2)$ $ ga (g^ {-1}g)bg^ {-1}=g (ab)g^ {-1}$ I'm stuck at this point, Is it correct so far? is ...
With all due respect to Manton, if that is indeed what he does, defining the map from $\mathrm {SU}_2$ to $\mathrm {SO}_3$ by this formula is a god-awful way to describe the map! Any element of $\mathrm {U}_2$ is a matrix of the form $$ 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 ...
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 ...