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  • Reflexive Generalized Inverse - Mathematics Stack Exchange
    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)
  • Let $G$ be a group, $a \\in G$. Prove that for all $g \\in G$, $|a . . .
    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 person who answered your question You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?
  • Conjugacy Classes of the Quaternion Group $Q$
    Disclaimer: This is not exactly an explanation, but a relevant attempt at understanding conjugates and conjugate classes
  • Functional equation $x^2 (f (x)+f (y))= (x+y)f (yf (x))$ [closed]
    I think there's an interesting math question from a new user here, but unfortunately it was received much worse than it deserved @Jack costaux You can edit the question to improve it In particular: (1) Your original question asked how to find a solution of this equation If it stays same, I suggest to remove the solution-verification tag and the part "if i made any mistake clarify", so that
  • Let $a \\in G$. Show that for any $g \\in G$, $gC(a)g^{-1} = C(gag^{-1})$.
    Try checking if the element $ghg^ {-1}$ you thought of is in $C (gag^ {-1})$ and then vice versa
  • Prove that $o (a)=o (gag^ {-1})$ - Mathematics Stack Exchange
    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
  • Subgroup that contains all Sylow $p$-subgroups
    You could "copy" the proof of Frattini's Argument: let $\;g\in G\;$ and let $\;P\;$ be some Sylow subgroup of $\;G\;$ , so $\;P\le H\;$ But also $\;P^g:=g^ {-1}Pg\;$ is a Sylow subgroup and thus also $\;P^g\le H $ Note that all the Sylow subgroups of $\;G\;$ are also Sylow subgps of $\;H\;$ , so by Sylow's Theorems $$\exists\;h\in H\;\;s t \;\;P^g=P^h\implies P^ {gh^ {-1}}=P\implies gh
  • Centralizer : center subgroup ~ normalizer : normal subgroup . . .
    The normalizer of 𝐴 in 𝐺 is the largest subgroup of 𝐺 that contains 𝐴 and in which 𝐴 is normal -> in which 𝐴 is normal to the normalizer correct?
  • abstract algebra - Show that conjugate by $g$ is isomorphism . . .
    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 $\sigma_ {g^ {-1}}\circ \sigma_ {g}=\sigma_ {g}\circ \sigma_ {g^ {-1}}=\operatorname {id}_G$ - And a map with left and right inverse map is bijective Then again, this does not reall ydiffer from what you wrote, does it?
  • Quotient group of a semidirect product. - Mathematics Stack Exchange
    Since $B\lhd G$, then $gbg^ {-1}\in B$, but what can I say about $gag^ {-1}$? I need to be able to say that $gag^ {-1}\in A$ to complete this part of the proof, right?





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