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).
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 ...
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.
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 $N_G (A)$!
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 $\gamma_ {g^ {-1}}=a\mapsto g^ {-1}ag$.
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