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 …

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 ⊆ G A ⊆ G is the set of all g ∈ G g ∈ G such that gAg−1 = A g A g 1 = A — which is exactly the normalizer …

Sep 20, 2015 · Now (gag−1)2 = ⋯ = e (g a g 1) 2 = = e. So if a a is the only element of order 2 then a = gag−1 a = g a g 1, or ag = ga a g = g a for all g ∈ G g ∈ G.

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

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 …

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 …

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 …

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^ { …

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 ...