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 …

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.

Dec 5, 2018 · Try checking if the element $ghg^ {-1}$ you thought of is in $C (gag^ {-1})$ and then vice versa.

Disclaimer: This is not exactly an explanation, but a relevant attempt at understanding conjugates and conjugate classes.

That implies $A \subseteq gAg^ {-1}$ for any $g \in G$, which is the same as $g^ {-1}Ag \subseteq A$ for any $g \in G$ and this proves the normality of $A$ in $G$.

Jul 1, 2016 · I am trying to prove that $gAg^ {-1} \subset A$ implies $gAg^ {-1} = A$, where A is a subset of some group G, and g is a group element of G. This is stated without proof in Dummit and Foote.

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 …

Dec 7, 2011 · We have a group $G$ where $a$ is an element of $G$. Then we have a set $Z (a) = \ {g\in G : ga = ag\}$ called the centralizer of $a$. If I have an $x\in Z (a)$, how ...

Oct 18, 2015 · 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^ { …

Apr 14, 2023 · After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the …