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Sep
9
revised Is there a simple group of order $105$?
added 246 characters in body
Sep
9
comment Is there a simple group of order $105$?
@Joseph: Yes, sorry for for misleading remark. You-sir-33433 has given a better reason.
Sep
9
answered Is there a simple group of order $105$?
Sep
9
revised Is there a simple group of order $105$?
cleaned up the post
Sep
1
comment Every non-empty subset of the integers which is bounded above has a largest element.
Hmmm, maybe you are right. Perhaps it is better to emphasize that $S$ (being as a subset of $\mathbb{Z}$) is also a subset of $\mathbb{R}$. So relational operator $<$ makes sense on $S$. I am still not sure if we have pinpointed the exact problem. @k73586: Please feel free to elaborate! :)
Sep
1
comment An example of set with a countably infinite set of accumulation points
@user162343: Yes, exactly. The union $S$ has countably infinite set of accumulation points: Indeed, $\{0, 1, 2, 3, …\}$ is the set of accumulation points for the set $S$.
Sep
1
answered Every non-empty subset of the integers which is bounded above has a largest element.
Sep
1
answered An example of set with a countably infinite set of accumulation points
Aug
31
comment Enlightening proof that the algebraic numbers form a field
Beautiful proof.
Aug
27
accepted $\int f = \lim\int f$ but $\int_{E}f\neq\lim\int_{E} f_{n}$
Aug
27
comment $\int f = \lim\int f$ but $\int_{E}f\neq\lim\int_{E} f_{n}$
Clever :) I like it. Thanks a lot!
Aug
27
asked $\int f = \lim\int f$ but $\int_{E}f\neq\lim\int_{E} f_{n}$
Aug
15
comment Show that $\mathbb Z[x]$ and $\mathbb Q_{>0}$ are isomorphic
Perhaps this could be of help: Think of elements of $(\mathbb{Z}(x), +)$ as infinite tuples which are eventually zero. Explicitly, you can identify $a_{0}+a_{1}x+a_{2}x^2+\cdots + a_{n} x^{n}$ with the infinite tuple $(a_{0}, a_{1}, …, a_{n}, 0, 0, …)$ where $a_{i}\in\mathbb{Z}$. Can you think of a map that would send this element to $(\mathbb{Q}_{>}, \cdot)$? (Another hint is to consider uniqueness of prime factorization).
Aug
15
reviewed Approve suggested edit on Show that $\mathbb Z[x]$ and $\mathbb Q_{>0}$ are isomorphic
Aug
14
answered Understanding what the Sylow theorems say about $p$-groups
Aug
14
comment Deduction of usual Cayley-Hamilton Theorem from “Determinant Trick”
@SammyBlack: Hmmm, why does it make sense only for those endomorphisms acting on the right?
Aug
14
awarded  Custodian
Aug
14
asked Deduction of usual Cayley-Hamilton Theorem from “Determinant Trick”
Aug
13
comment Normalizer and centralizer of abelian subgroups of a group are equal
@user1729: No worries! :)
Aug
13
comment Normalizer and centralizer of abelian subgroups of a group are equal
@kammy: Yes. Using $H=Z(G)$ only gives $G=G$.