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 …

We have a group G G where a a is an element of G G. Then we have a set Z(a) = {g ∈ G: ga = ag} Z (a) = {g ∈ G: g a = a g} called the centralizer of a a. If I have an x ∈ Z(a) x ∈ Z (a), how do I go about …

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.

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 ...

Jan 3, 2019 · 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 $N_G (A)$!

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

Sep 27, 2015 · 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 …

Jul 9, 2015 · $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 ...

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.