gag gifts under $5 - Gadget World
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. Try checking if the element $ghg^ {-1}$ you thought of is in $C (gag^ {-1})$ and then vice versa. Let $a \in G$.
Understanding the Context
Show that for any $g \in G$, $gC (a)g^ {-1} = C (gag ... 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 $\gamma_ {g^ {-1}}=a\mapsto g^ {-1}ag$. 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).
Image Gallery
Key Insights
$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 ... More generally, the one element subset $\ {a\}$ is a conjugacy class if and only if $gag^ {-1} = a$ for all $g \in G$ if and only if $a$ is in the center of $G$. 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 $N_G (A)$!