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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
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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
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comment Showing $A_5$ is simple
Yes, that is exactly what I mean, so that the counterexample holds.
Nov
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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
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comment Showing $A_5$ is simple
Yes, but it does not divide the order of $A_5/H$.
Nov
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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
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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.
Nov
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comment Proving that f is measurable if g is measurable where $g(x, y) = |f(x) - f(y)|$
Ok, then, do you still have doubts?
Nov
11
comment Proving that f is measurable if g is measurable where $g(x, y) = |f(x) - f(y)|$
Ah, my definition of integrability for an arbitrary real-valued function is that $f$ is integrable if and only if $|f|$ is integrable. You have a different one?
Nov
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comment Proving that f is measurable if g is measurable where $g(x, y) = |f(x) - f(y)|$
Yeah, you're right!
Nov
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answered Finding $E(XY)$ given $E(X)$, $E(Y)$, $E(Y^2)$, $E(X|Y)$
Nov
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answered Proving that f is measurable if g is measurable where $g(x, y) = |f(x) - f(y)|$
Nov
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answered Showing that a measure is lower continuous.
Nov
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comment Proof of Gauss lemma
You're welcome!