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Apr
21
comment When does each action corresponds to a homomorphism and an anti-homomorphism.
Yes. Thank you, I meant left but wrote right (I edited it), if we have a right action, then $\varphi_g(x) := xg$ gives an anti-homomorphism, as $\varphi_{gh}(x) = x(gh) = (xg)h = \varphi_h(xg) = \varphi_h(\varphi_g(x)) = (\varphi_h \circ \varphi_g)(x)$, but for a left action the definition $\varphi_g(x) := gx$ gives a homomorphism w.r.t. the way I write function application and composition.
Apr
21
revised When does each action corresponds to a homomorphism and an anti-homomorphism.
deleted 1 character in body
Apr
21
asked When does each action corresponds to a homomorphism and an anti-homomorphism.
Apr
19
awarded  Popular Question
Apr
19
answered Is the following a valid characterisation of complete metric spaces?
Apr
19
asked Is the following a valid characterisation of complete metric spaces?
Apr
10
comment Components and Centralisers of Involution. A wrong argument?
Yes, thank you! Don't know why I was blind for this...
Apr
10
asked Components and Centralisers of Involution. A wrong argument?
Apr
9
comment What does it mean for a subgroup to be self-centralising in terms of group extensions
Ah guess I found it out. For a given Fitting subgroup $F$ there are just a finite number of subgroups $U$ which contain the center $Z(F)$, and also $\mbox{Aut}(F)$ is finite, and if $V \le \mbox{Aut}(F)$ also there are just a finite number of homomorphisms $\varphi : U \to \mbox{Aut}(V)$, so there are just a finite number of solvable groups that could be constructed as a semidirect product $G = V \ltimes_{\varphi} U$, in particular so that $U = C_G(F)$, in this way we have a bound on the number of groups that are possible. And that might mean "the Fitting subgroup controls the structure".
Apr
9
accepted Sufficient Conditions for the Commutator Subgroup to be a Component
Apr
9
revised Sufficient Conditions for the Commutator Subgroup to be a Component
edited body
Apr
9
asked Sufficient Conditions for the Commutator Subgroup to be a Component
Apr
9
comment What does it mean for a subgroup to be self-centralising in terms of group extensions
Thanks for your answer! And thats everything behind the phrase "the Fitting subgroup controls the structure of a solvable group."?
Apr
9
asked What does it mean for a subgroup to be self-centralising in terms of group extensions
Apr
9
comment Subgroups and Direct Products of $\pi$-closed groups are also $\pi$-closed.
@j.p. Let $X := p_1(O_{\pi}(G\times H)), Y := p_2(O_{\pi}(G\times H))$, then as $X,Y$ are normal and $\pi$-groups, we have $X \le O_{\pi}(G), Y \le O_{\pi}(H)$ and therefore $X\times Y \le O_{\pi}(G)\times O_{\pi}(H)$. Also for arbitrary sets $Z \le A\times B$ we have $Z \le p_1(Z)\times p_2(Z)$, and therefore $O_{\pi}(G\times H) \le X\times Y \le O_{\pi}(G)\times O_{\pi}(H)$, so this gives the other direction. Thanks!
Apr
9
comment Subgroups and Direct Products of $\pi$-closed groups are also $\pi$-closed.
@j.p. We have $O_{\pi}(G) = O_{\pi}(G\times H) \cap G$, but I do not see that $O_{\pi}(G \times H) \le (O_{\pi}(G\times H)\cap G)\times (O_{\pi}(G\times H)\cap H)$ has to hold, as in general the relation $U \le (U\cap A)\times (U\cap B)$ does not hold, for example take $U = \{ (0,0), (1,1) \}$ and $A=B=C_2$.
Apr
8
comment Subgroups and Direct Products of $\pi$-closed groups are also $\pi$-closed.
@DerekHolt: For me $O_{\pi}(G)\times O_{\pi}(H) \le O_{\pi}(G\times H)$ is obvious, but $O_{\pi}(G\times H) \le O_{\pi}(H)\times O_{\pi}(G)$ not so?
Apr
8
asked Subgroups and Direct Products of $\pi$-closed groups are also $\pi$-closed.
Apr
7
comment Proofing facts about Factor groups and homomorphic images by direct computation rather then using the isomorphism theorems
Can you please leave a short comment when you downvote why you think so...
Apr
7
answered Can the Quotient Group $HK/K$ be Confusing?