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1d
accepted For $N, M \unlhd G$ relation between $MN/(M\cap N)$ and $N/(M\cap N)\times M/(M\cap N)$
1d
asked For $N, M \unlhd G$ relation between $MN/(M\cap N)$ and $N/(M\cap N)\times M/(M\cap N)$
1d
comment If two quotient groups are semi-simple, then a third build from both is semi-simple too.
Thanks for your contribution. But as you wrote in your first paragraph $N \le A_i$ so we have $A_i \vee N = A_i$, which renders your first case impossible and so it is not necessary to consider it, or have I overlooked something?
1d
comment Inequality for the set of factors of length $n$ of some regular language
@wece The minimal automaton accepting a given language is unique up to isomorphism, this in particular implies that for a minimal automaton the number of states is unique too.
Aug
20
comment Inequality for the set of factors of length $n$ of some regular language
Yes, of course. But the minimal one is unique.
Aug
15
accepted Group $G$ acting on $\Omega$ such that each $\alpha \in \Omega$ has unique $p$-element fixing $\alpha$.
Aug
15
comment Group $G$ acting on $\Omega$ such that each $\alpha \in \Omega$ has unique $p$-element fixing $\alpha$.
... ohh its just the $p$-group fixed point theorem again. Set $X = \langle x \rangle$, which is a $p$-group and $\Delta^X = \Delta$, hence $|\mbox{fix}(X)| \equiv |\Delta| \pmod{p}$, so that $|\mbox{fix}(X)| \equiv 1 \pmod{p}$. So we have points in $\Delta$ fixed by all elements from $X$, in particular there exists some point fixed by $x$.
Aug
15
comment Group $G$ acting on $\Omega$ such that each $\alpha \in \Omega$ has unique $p$-element fixing $\alpha$.
Okay, I guess I got the first one. If $\Delta$ is a finite orbit, choose some $\alpha \in \Delta$. Then $\Delta^{Q(\alpha)} = \Delta$. So by the $p$-group fixed point theorem we have $|\mbox{fix}(Q(\alpha))| \equiv |\Delta| \pmod{p}$, and with $\mbox{fix}(Q(\alpha)) = \{\alpha\}$ the first congruence follows. For the other assertion: Any $p$-element $x$ such that $\Delta^x = \Delta$ with $|\Delta| \equiv 1 \pmod{p}$ has to fix some element from $\Delta$ I have a feeling that it must be right, but I cannot formulate a formal argument for it...
Aug
15
comment Group $G$ acting on $\Omega$ such that each $\alpha \in \Omega$ has unique $p$-element fixing $\alpha$.
I understand the finite case, but your last line I do not get, why do we have $|\Delta| \equiv 1 \pmod{p}$ and why must any $p$-element fix some point in $\Delta$?
Aug
15
revised Group $G$ acting on $\Omega$ such that each $\alpha \in \Omega$ has unique $p$-element fixing $\alpha$.
deleted 1141 characters in body
Aug
15
comment Group $G$ acting on $\Omega$ such that each $\alpha \in \Omega$ has unique $p$-element fixing $\alpha$.
Yes. Thank you for pointing out, somehow I thought the $p$-element is unique, but of course this is not given by the exercise.
Aug
15
revised Group $G$ acting on $\Omega$ such that each $\alpha \in \Omega$ has unique $p$-element fixing $\alpha$.
added 104 characters in body
Aug
15
asked Group $G$ acting on $\Omega$ such that each $\alpha \in \Omega$ has unique $p$-element fixing $\alpha$.
Aug
14
accepted Formal proof of a simple fact, namely that $S$ has even cardinality if certain pairs could idenitifed
Aug
14
comment Formal proof of a simple fact, namely that $S$ has even cardinality if certain pairs could idenitifed
You all gave great answers, and I really like all of them! But as I need to select one, I choose this for its conciseness and somewhat unusual construction.
Aug
14
comment Formal proof of a simple fact, namely that $S$ has even cardinality if certain pairs could idenitifed
For case (2). Do you forgot to define $R_k$? Guess it must be $R_k := R_{k-1} \setminus\{x, f(x)\}$.
Aug
14
asked Formal proof of a simple fact, namely that $S$ has even cardinality if certain pairs could idenitifed
Aug
14
accepted Must a solution of $x^2 + 4y^2 = p$ be unique
Aug
14
comment Must a solution of $x^2 + 4y^2 = p$ be unique
Thank you for your help, would you mind writing an answer so I can check this question as answered...
Aug
14
comment Must a solution of $x^2 + 4y^2 = p$ be unique
I heard of them. Okay with the help of wikipedia I can see the arguments: we have $x^2 + 4y^2 = (x - 2iy)(x+2iy)$ and both $x-2iy$ and $x+2iy$ are primes in $\mathbb Z[i]$, which implies that this factorisation is unique (up to units), i.e. $x$ and $y$ are unique too, right?