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location Mexico City
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visits member for 2 years, 6 months
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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.


4m
comment How to solve this kind of problem?
The number in the bottom right corner has to be $1,-1,11$ or $-11$
15m
answered Check if Sequence is Graphic: 8, 8, 7, 7, 6, 6, 4, 3, 2, 1, 1, 1
1h
comment Is it possible to permute an unknown binary sequence so that two particular bits are equal?
I editted it, is it convincing now?
1h
revised Is it possible to permute an unknown binary sequence so that two particular bits are equal?
added 886 characters in body
13h
awarded  abstract-algebra
18h
comment An abelian subgroup of symmetric group
Wow, that was pretty sweet. Using abelians are clt was rather ingenious.
19h
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.
20h
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.
20h
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?
21h
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 $r<k$. Because the sequence will be the same. (Of course it is possible the swap didn't actually occur, but for all intents and purposes we can assume it did, since the outcome would be the same).
21h
comment Is it possible to permute an unknown binary sequence so that two particular bits are equal?
By induction the sequence will always be a permutation of the $b$'s. So if after a number of steps he can determine the bit at position $m$ is the same as the one in position $n$ he knows that those bits correspond to $b_k$ and $b_r$ for example(wlog $k<r$). and then he knows that the number of zeroes is not $k$ (since otherwise they would be different).
21h
comment Is it possible to permute an unknown binary sequence so that two particular bits are equal?
Hmm, what I mean is the following: The mathematician starts with the ordered sequence $b_1,b_2,b_3\dots b_{2015}$. In the first move if he sends an electron from $b_r$ to $b_k$ with $r<k$ he knows that the sequence will stay the same. so the sequence is still $b_1,b_2\dots b_{2015}$. if $r>k$ 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.
21h
accepted Is it possible to permute an unknown binary sequence so that two particular bits are equal?
21h
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.
21h
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.
21h
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 $r<k$ we know that nothing will happen. And on the other hand, when $r>k$ we can safely assume the value $b_r$ has been switched with the value of $b_k$
21h
revised Is it possible to permute an unknown binary sequence so that two particular bits are equal?
added 98 characters in body
21h
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.
21h
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.
21h
answered Is it possible to permute an unknown binary sequence so that two particular bits are equal?