# Jorge Fernández

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bio website location Mexico City age member for 2 years, 6 months seen 4 hours ago profile views 1,957

Hello, I am a freshman at The National Autonomous University of Mexico (UNAM).

My interests are contest problems,elementary number theory, combinatorics, graph theory, linear algebra, group theory, abstract algebra and I have been meaning to learn category theory and matroids.

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 3h awarded abstract-algebra 9h comment An abelian subgroup of symmetric group Wow, that was pretty sweet. Using abelians are clt was rather ingenious. 10h comment Normal subgroups in groups of odd order I don't see why $H$ needs to be normal for the first proof. Couldn't we use the same argument without normality?That is: the order of $N_G(H)/C_G(H)$ divides the order of $N_G(H)$ and therefore the order of $G$. But it also divides the order of $Aut(H)$? since the order of $Aut(H)$ is relatively prime to the order of $G$ we get$C_G(N)=N_G(N)$. Oh, I see why you need it to be normal lol. 10h comment Normal subgroups in groups of odd order proposition $2$ can be proven using the thing about normal subgroups being a union of conjugacy classes.the smallest non trivial ones have size $p$. Although prop 2 is not generalization. 10h comment Normal subgroups in groups of odd order The proof of the proposition is the same isn't it? You get the order of $N_G(H)/C_G(H)$ divides the order of $\Aut(H)=p-1$ and the order of $N(G)$. Why does $H$ have to be normal though? Can't you get the same results without $H$ being normal? 11h comment Is it possible to permute an unknown binary sequence so that two particular bits are equal? My point is that he can safely assume the swap was done when $r>k$ and he can safely assume the swap was not done when $rk$ he knows that the sequence is going to be the same as the sequence when $b_r$ and $b_k$ are switched (because the only possibilities are $b_r=b_k=1$ or $b_r=1,b_k=0$. In both cases the string becomes the string when $b_r$ is switched with $b_k$ and everything stays the same. 11h accepted Is it possible to permute an unknown binary sequence so that two particular bits are equal? 11h comment Is it possible to permute an unknown binary sequence so that two particular bits are equal? well yes, but in this case the mathematician knows which where the "coordinates" of the diodes he did, what I mean is that he can reverse his own process. 11h comment Is it possible to permute an unknown binary sequence so that two particular bits are equal? I posted a solution, could you comment please? thank you. 11h comment Is it possible to permute an unknown binary sequence so that two particular bits are equal? When I say that we can fully expect the outcome I mean that when we send an electron from $b_r$ to $b_k$ with $rk$ we can safely assume the value $b_r$ has been switched with the value of $b_k$ 12h revised Is it possible to permute an unknown binary sequence so that two particular bits are equal? added 98 characters in body 12h comment Is it possible to permute an unknown binary sequence so that two particular bits are equal? Note that in general we can't reverse the process, but when they are ordered at first we can do it. 12h comment Is it possible to permute an unknown binary sequence so that two particular bits are equal? Thanks, although the original problem is doable if we know there are at least two $0$ digits. 12h answered Is it possible to permute an unknown binary sequence so that two particular bits are equal? 12h comment combination and permutation SETS ok, then you need to select a non empty subset of the 11 graders, a non empty subset of the 12 graders and a subset of all the others. 12h comment Is it possible to permute an unknown binary sequence so that two particular bits are equal? yeah, that's what I thoug, although I seem to have copied the problem wrong. 12h comment Is it possible to permute an unknown binary sequence so that two particular bits are equal? Oh, you are right. 12h comment Is it possible to permute an unknown binary sequence so that two particular bits are equal? swapping to equal bits is the same as not swapping them for all intents and purposes.