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22h
comment If $G$ and $G'$ are two finite group of same cardinal, then $G\cong G'$.
I think we also need $pq$ where $p<q$ and $q$ is not congruent to $1\bmod p$.
22h
comment If $G$ and $G'$ are two finite group of same cardinal, then $G\cong G'$.
Oh, you seem to be right.
22h
comment If $G$ and $G'$ are two finite group of same cardinal, then $G\cong G'$.
I am pretty sure there are at least two non-isomorphic groups of any cardinality except for $0,1$ and $p$ (where $p$ is prime).
23h
revised Finding a closed formula for: $1\cdot2\cdot3+2\cdot3\cdot4+…+(n-2)\cdot(n-1)\cdot(n)$
deleted 2 characters in body
23h
answered Finding a closed formula for: $1\cdot2\cdot3+2\cdot3\cdot4+…+(n-2)\cdot(n-1)\cdot(n)$
1d
comment How do I prove that if $2\nmid n$ then $2|(n+1)$?
what can you use?
1d
comment What is the probability that a psychic correctly “predicts” the outcome of at least 5 out of 10 coin flips?
yes${}{}{}{}{}$
2d
answered What is the probability that a psychic correctly “predicts” the outcome of at least 5 out of 10 coin flips?
2d
revised Smallest $n$ such that $U(n)$ contains a subgroup isomorphic to $\mathbb Z_5 \oplus \mathbb Z_5$
added 707 characters in body
2d
answered Smallest $n$ such that $U(n)$ contains a subgroup isomorphic to $\mathbb Z_5 \oplus \mathbb Z_5$
2d
comment How I can prove that for any natural number $n$ such that $30<n$, $\pi(4n-3)<n$?
it is true, we have $\pi(x)<1.3\frac{x}{\log x}$ for $x\leq 17$
2d
answered Is $S_1\cap S_2$ and $S_1\setminus S_2$ always linearly dependent if $S_1$ and $S_2$ are linearly dependent subsets of vector space $V$?
2d
comment Show that $1^k+2^k+\cdots+n^k$ is $\Omega (n^{k+1})$
because $1^k+2^k+3^k+\dots+n^k=\int\limits_0^n\lceil x \rceil ^kdx\leq\int\limits_0^n x ^kdx$
Feb
7
revised Show that $1^k+2^k+\cdots+n^k$ is $\Omega (n^{k+1})$
added 150 characters in body
Feb
7
revised Show that $1^k+2^k+\cdots+n^k$ is $\Omega (n^{k+1})$
added 150 characters in body
Feb
7
revised Show that $1^k+2^k+\cdots+n^k$ is $\Omega (n^{k+1})$
added 4 characters in body
Feb
7
answered Show that $1^k+2^k+\cdots+n^k$ is $\Omega (n^{k+1})$
Feb
7
answered Show that an inverse of a bijective linear map is a linear map.
Feb
7
comment Is there an arbitrarily large set of naturals so that the sum of each two has exactly $n$ prime divisors? What about an infinite set?
exactly $n$ prime divisors right?
Feb
7
awarded  Nice Answer