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| location | ||
| age | ||
| visits | member for | 1 year, 10 months |
| seen | 18 mins ago | |
| stats | profile views | 91 |
Computer science and engineering undergraduate & mathematics enthusiast.
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Jun 15 |
accepted | Atiyah-Macdonald, Proposition 2.12, uniqueness of the tensor product. |
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Jun 15 |
comment |
Atiyah-Macdonald, Proposition 2.12, uniqueness of the tensor product. Oh, that was indeed very simple. Thanks. |
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Jun 15 |
asked | Atiyah-Macdonald, Proposition 2.12, uniqueness of the tensor product. |
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May 10 |
awarded | Nice Question |
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Aug 16 |
awarded | Yearling |
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Jul 11 |
comment |
Exercise on modules over PID involving injective modules, Baer's criterion. @navigetor Exercise is from Hungerford. It is self-study, I complement it with Robert Ash's text Basic Abstract Algebra. |
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Jul 11 |
comment |
Exercise on modules over PID involving injective modules, Baer's criterion. Thanks for reading it. I sort of thought it was too long for anybody to care enough to get through it. Indeed author leads the reader into a solution, but I need some confirmation I somewhat understood the material correctly, as similar concepts occur frequently later in the book. |
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Jul 10 |
revised |
Exercise on modules over PID involving injective modules, Baer's criterion. deleted 39 characters in body |
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Jul 9 |
asked | Exercise on modules over PID involving injective modules, Baer's criterion. |
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Jun 8 |
awarded | Caucus |
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Apr 3 |
accepted | Infinite nilpotent group, any normal subgroup intersects the center nontrivially |
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Apr 2 |
revised |
Infinite nilpotent group, any normal subgroup intersects the center nontrivially deleted 2725 characters in body |
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Apr 2 |
answered | Infinite nilpotent group, any normal subgroup intersects the center nontrivially |
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Apr 2 |
revised |
Infinite nilpotent group, any normal subgroup intersects the center nontrivially added 1141 characters in body |
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Apr 2 |
comment |
Infinite nilpotent group, any normal subgroup intersects the center nontrivially @Steve Indeed, this proof is much cleaner. I am not completely comfortable I understand it right, but actually Hungerford characterises nilpotent groups in a similar way in one of the exercises (nilpotent iff $\gamma_n(G) = \langle e \rangle$ for some $n$, where $\gamma_1(G) = G, \gamma_2(G) = (G,G), \gamma_i(G) = (\gamma_{i-1}(G), G)$. I will have to look into this, but is it correct that $N_i \subset \gamma_i(G)$ ? |
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Apr 1 |
revised |
Infinite nilpotent group, any normal subgroup intersects the center nontrivially added 33 characters in body |
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Apr 1 |
revised |
Infinite nilpotent group, any normal subgroup intersects the center nontrivially added 1655 characters in body |
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Apr 1 |
comment |
Infinite nilpotent group, any normal subgroup intersects the center nontrivially @Ted oh, of course. In the quotient group $G/Z_i$ the subgroup $Z_{i+1}/Z_i$ is the center. But that implies for any $c \in Z_{i+1}$, $Z_icZ_ig = Z_icg = Z_igc$, and therefore $Z_i = Z_igcg^{-1}c^{-1}$ which means $gcg^{-1}c^{-1} \in Z_i$ for all $g \in G, c \in Z_{i+1}$. So the all generators of $[G,Z_{i+1}]$ are contained in $Z_i$ and therefore the entire group is too. I think I should be able to get the rest of the proof from here, and will answer my own question later today or tomorrow, unless somebody beats me to it. Big thanks for everybody! |
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Apr 1 |
comment |
Infinite nilpotent group, any normal subgroup intersects the center nontrivially @alex, please excuse my ignorance. I'm reading the solution in the link. Why is $[G, Z_{i+1}(G)] \subset Z_i(G)$ ? I see that commutator of $Z_{i+1}(G)$ must be contained in $Z_{i}(G)$ as quotient $Z_{i+1}(G)/Z_{i}(G)$ is abelian, but why the entire $[G,Z_{i+1}(G)]$ ? |
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Apr 1 |
accepted | Elementary field theory, field extensions of the rationals of degree 2 |
