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Souvik Dey
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 Mar20 comment Pull back image of maximal ideal under surjective ring homomorphism is maximal But $J \subseteq f^{-1}(M)$ is a contradiction because I assumed $f^{-1}(M)$ is a "proper subset" of $J$ ... Mar20 asked Pull back image of maximal ideal under surjective ring homomorphism is maximal Feb23 comment If $G$ is an uncountable group and $H$ is a subgroup then $G$ \ $H$ is uncountable ? @FrankScience: But how ?? a countable union of uncountable sets is also uncountable ... Feb23 comment If $G$ is an uncountable group and $H$ is a subgroup then $G$ \ $H$ is uncountable ? @FrankScience: Yes $G-H$ is indeed the set difference and $xH$ do have same cardinality as that of $H$ Feb23 asked If $G$ is an uncountable group and $H$ is a subgroup then $G$ \ $H$ is uncountable ? Feb22 accepted How to prove $(\space|\sin n|\space)$ does not converge? Feb8 answered Does $G\cong G/H$ imply that $H$ is trivial? Jan24 comment A simple way to obtain $\prod_{p\in\mathbb{P}}\frac{1}{1-p^{-s}}=\sum_{n=1}^{\infty}\frac{1}{n^s}$ @nullUser: You have shown $E_p$ 's are independent , but in the last line of the proof aren't you assuming $E^c_p$'s are independent ? Jan16 awarded Socratic Jan5 comment Finding partitions $\{A,B\}$ of the set $\mathbb N_n:=\{1,…,n\}$ such that $\Big|\prod_{i \in A}p_i-\prod_{i\in B}p_i\Big|=1$ @GerryMyerson : :) yes , I have seen your answer also upvoted it Jan5 comment Any ring of prime order commutative ? How can I get $x$ from $\bar x$ ? Jan3 comment Proving that $f(n)=n$ if $f(n+1)>f(f(n))$ @MikeSpivey: In the proof of claim 3 : from $f(f(k-1))0$ ? I seem to only conclude $h \ge 0$ .... Dec29 comment On integer $n>1$ and prime $p$ such that $p1$ and prime $p$ such that $p1$ , does there exist a group $G$ with elements $a,b \in G$ such that $o(a)=m , o(b)=n$ but $ab$ has infinite order ? Dec28 comment For any integers $m,n>1$ , does there exist a group $G$ with elements $a,b \in G$ such that $o(a)=m , o(b)=n$ but $ab$ has infinite order ? @mesel: but this doesn't assure we can always find a group with required elements for arbitrary integers $m,n >1$ ; there is a result stating "For any integers $m,n,r$, all greater than $1$ , there is a finite group $G$ with elements $a,b\in G$ such that $o(a)=m,o(b)=n,o(ab)=r$ perhaps you can see my motivation ...