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Apr
25
comment Prove that this set is open
@drhab - I think you should keep it, it can be interesting for future readers as it is a confusing matter
Apr
25
answered On convergent sequences
Apr
25
comment Prove that this set is open
@drhab - It is prove! If $X=\Pi_{i=1}^{n}\mathbb{R}$ one can consider two topologies on $X$ - The product topology and the topology induced by the metric on $\mathbb{R}^{p}$. I proved that $A$ was open without relying on the fact that the two topologies on $X$ are the same since I assumed the OP does not know that
Apr
25
answered Prove that this set is open
Apr
25
accepted Difference between definitions of $p$-subgroup and Sylow $p$-subgroup
Apr
24
asked Difference between definitions of $p$-subgroup and Sylow $p$-subgroup
Apr
24
comment Topology, maps, continuity
@smits - then $f^{-1}(U)$ is closed. This condition is equivalent for the continuity of $f$
Apr
24
answered Proving that $3^n<n!$ when $n\geq 7$
Apr
24
answered Where could (do?) we go after exhausting greek letters?
Apr
24
answered Eigen space of $T_{A}$ where $T_{A}(v)=Av$
Apr
24
answered Proving that if a set is both open and closed then it is equal to the real numbers
Apr
24
answered Classify $ \mathbb{Z}_9\times\mathbb{Z}_8\times\mathbb{Z}_8$/<(3,2,4)> according to the fundamental theorem of finitely generated abelian groups.
Apr
24
awarded  Notable Question
Apr
23
comment Subgroups of every order dividing the order of the group imply the group is abelian?
Thank you for your answer, Can you please rephrase or explain the theorem you are taking about "if divides the order of a group with a prime, then the group has a subgroup of order " it seems to be missing a word or two and it sounds like Cauchy theorem..
Apr
23
comment If $f^2$ integrable, then $|f|$ is integrable?
Don't you need to know that $|f|$ (or $f$) is integrate to begin with ? The space of which this inequality is deduced is the space of integrable functions $f$
Apr
23
accepted Subgroups of every order dividing the order of the group imply the group is abelian?
Apr
23
comment Subgroups of every order dividing the order of the group imply the group is abelian?
@Timbuc - I see your point. But this seems to answer only (2), can you also give an example for (1) ? thanks!
Apr
23
comment Groups with “few” subgroups
@mathmandan - Very interesting, I'll be sure to read it. thanks!
Apr
23
asked Subgroups of every order dividing the order of the group imply the group is abelian?
Apr
23
comment Groups with “few” subgroups
@mathmandan - How do you know such subgroups of $T$ exist ?