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16h
comment A group-ring is commutative if and only if that group is abelian
Note that since multiplication in your algebra is defined by the multiplication of the basis elements, that is $g\cdot h=gh$ where the RHS is multiplication in $G$, your algebra will be commutative iff the basis elements commute with themselves. This amounts to saying $G$ is commutative.
16h
answered How to show $\mathbb{R}^2/\mathbb{Z}^2$ is homeomorphic to $\mathbb{R}/\mathbb{Z} \times \mathbb{R}/\mathbb{Z}$
19h
comment Proving the cardinality of $|A| =|\mathbb Z|$
If something is enumerated by another set $I$ it has cardinality at most that of such set, for there is a surjection from $I$ to such enumerated set. Since in your case your set is at most countable and infinite, it is countably infinite.
1d
answered how to prove that $C^{k}$ map does not depend on choice of the charts
1d
comment How many ways are there to prove Cayley-Hamilton Theorem?
@NNN They are useful for future reference.
2d
comment Extending the automorphism of $Q(\sqrt2)$ to automorphism of $Q(\sqrt(1+\sqrt2))$.
Your bigger extension is not normal. It doesn't contain, in particular, the two other complex roots. ($\sqrt 2 >1$!). In fact, you cannot extend your morphism.
May
2
answered Why does $\sum_{k=1}^{\infty}\sum_{\ell=0}^{k-1} = \sum_{\ell=0}^{\infty}\sum_{k=\ell+1}^{\infty}$
May
2
comment Show that $S^1$ acts on $S^3$
@Epsilon Using $\bar z$ for $z^{-1}$ obscures the fact that if $z$ acts by an automorphism, then its inverse must be the automorphism by which $z^{-1}$ acts.
May
2
revised Part of proof of the set of continuous integrable functions is dense in $L^1(\Bbb R)$
edited title
May
1
comment If $(a_n)$ is positive and $\sum\limits_n \frac{a_n}{1+a_n}$ converges then $\sum\limits_n a_n$ converges
Adam Hughes' answer shows the hypothesis that $a_n>0$ is necessary, and gives $a_n = (-1)^n/n$ as a counterexample.
May
1
answered If $(a_n)$ is positive and $\sum\limits_n \frac{a_n}{1+a_n}$ converges then $\sum\limits_n a_n$ converges
May
1
comment Showing epimorphism without using the Freyd-Mitchell Embedding Theorem
What about using MacLane's technique of "elements"?
May
1
awarded  Popular Question
Apr
30
comment Part of proof of term-by-term integration
Keep looking! ${}$
Apr
30
comment Part of proof of term-by-term integration
This is proved in every book covering Lebesgue integration. Did you check those?
Apr
30
comment Catalan numbers and triangulation
It should be noted that the core of this argument is how to obtain a triangulation of a smaller n-gon given one of a larger one, and remembering how the collapsing was done gives the bijection. One has to track things carefully, is all.
Apr
30
comment Catalan numbers and triangulation
(This is not a combinatorial proof!)
Apr
30
comment Example of a diffeomorphism from all of $\mathbb{R}$ to itself
Any linear map, any affine transformation. Multiplication by a nonzero real.
Apr
30
comment Examples of Manifolds such that $\chi (X)=-3$
Try higher dimensions.
Apr
29
comment Let $f:[0,1] \rightarrow \mathbb{R}$ be a continuous function.Show that $\lim_{n \rightarrow \infty} n \int_{0}^{1}e^{-nx}f(x)dx=f(0)$
Evaluate the integral. $f(0)$ is constant.