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 …

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 ...

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.

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 …

Sep 7, 2024 · This is an exercise in Weibel "Homological Algebra", chapter 6 on group cohomology. For reference, this is on Page 183. So the question was asking us to ...

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 $G$.

Feb 24, 2020 · Prove that the relation $a\sim b$ if $b=gag^ {-1}$ for some $g\in G$, is an equivalence relation on $G$. Prove that $\forall u,v\in G$, $uv\sim vu$. So I've proved (1). My …

Jul 18, 2020 · Additionally, I would include an extra step in your commutation computation for clarity: \begin {align*} (gag^ {-1}) (hah^ {-1})&=ga (g^ {-1}hah^ {-1}g)g^ {-1}\\ &= ga\left ( (g^ { …