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seen Dec 16 at 18:02

Dec
15
comment Flexes of cubic curve
@evinda: I wrote an answer that hopefully explains enough. Sorry that it took me so long and tell me if you would like further clarifications.
Dec
15
answered Flexes of cubic curve
Dec
9
comment Flexes of cubic curve
sorry I have to go now, I will write a more extensive explanation later. Anyway you can try to have a look at wikipedia: en.wikipedia.org/wiki/Projective_space
Dec
9
comment Flexes of cubic curve
No, by definition $\mathbb{P}^2(\mathbb{C})=(\mathbb{C}^2\setminus \{0\}) / \mathbb{C}^*$.
Dec
9
comment Flexes of cubic curve
What is your definition of nonsingular? In this case you can just check that the partial derivatives of $F$ cannot vanish all together in a point of the curve.
Dec
9
comment Flexes of cubic curve
First you could try to prove that the curve is nonsingular, and then compute the interesection points of the curve with the Hessian. In this case the computations should be pretty straighforward.
Dec
4
answered Vanishing of global sections with very negative twists.
Nov
11
awarded  Yearling
Nov
11
comment Showing $A_5$ is simple
Yes, usually by $A_5/H$ you mean the set of all cosets of $H$, and if $H$ is normal you can give this the structure of a group and call it the quotient group (or also the factor group I guess, even if I never heard that name).
Nov
11
comment Showing $A_5$ is simple
If you want a proof that $A_5$ is simple then you can look at it here: people.math.gatech.edu/~ecroot/A5.pdf. However I think that you should check again your sources, because what you have seen is not true: $A_5$ is actually a coset of $H$, since $H=A_5$ and $A_5=eA_5$.
Nov
11
comment Showing $A_5$ is simple
Yes, that is exactly what I mean, so that the counterexample holds.
Nov
11
comment Showing $A_5$ is simple
Sorry, it seems that I have some problems with the chat. However, the coset $aH$ has the same order of $H$ so that $|aH|=|H|$. Then, what do you mean by |A_5/H| and the order of |a|? If you mean the cardinality of $A_5/H$ and the order of the element $a$, then taking $H=A_5$ and $a=id$ gives a counterexample. Indeed, if $|a|=1$,$|H|=60$ and $|A_5/H|=1$. Anyway if you want a proof there is a nicely explained one here:people.math.gatech.edu/~ecroot/A5.pdf
Nov
11
comment Showing $A_5$ is simple
Yes, but it does not divide the order of $A_5/H$.
Nov
11
comment Showing $A_5$ is simple
The $H$ in my example is normal as well, since $G$ is abelian and every subgroup of an abelian group is normal. Also, if in your case you consider $H=A_5$ then this is normal, but $|A_5|$ does not divide $|A_5/A_5|=1$. But if you are interested I am writing a proof of the simplicity in an answer.
Nov
11
comment Showing $A_5$ is simple
Yes it is true that $|H|$ divides $|A_5|$ but this does not guarantee that it divides also $|A_5/H|=\frac{|A_5|}{|H|}$. For example consider the group $G=\mathbb{Z}_2\times \mathbb{Z}_3$ and the subgroup $H=\mathbb{Z}_2\times {0}$. Then $|H|=2$ and this does not divide $|G/H|=3$. Also, taking $a=0$, you can see that this does not divide $|a|=1$ as well.
Nov
11
comment Showing $A_5$ is simple
Then the cardinality of $aH$ is the same as the cardinality of $H$ and in general this does not divide the order of $a$. Also, I cannot see why this divides $|A_5/H|$.
Nov
11
comment Showing $A_5$ is simple
What do you mean by $aH$ ?
Nov
11
comment Show f(x):=sqrt(x) is uniformly continuous on [0,1]
I think your solution is fine. In general, you can also invoke the Heine-Cantor Theorem.
Nov
11
comment Proving that f is measurable if g is measurable where $g(x, y) = |f(x) - f(y)|$
Sure: the inequality $|f(y)-f(x)|\leq |f(x)|+|f(y)|$ is the triangle inequality, and the other is still the triangle inequality applied to $f(y)=[f(y)-f(x)]+f(x)$.
Nov
11
comment Proving that f is measurable if g is measurable where $g(x, y) = |f(x) - f(y)|$
You're welcome. Happy that I could help.