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| bio | website | home.uchicago.edu/~/… |
|---|---|---|
| location | Chicago, IL | |
| age | 21 | |
| visits | member for | 4 months |
| seen | 2 days ago | |
| stats | profile views | 98 |
I am currently a 3rd year undergraduate student in the Department of Mathematics of the University of Chicago.
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May 6 |
awarded | Caucus |
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May 2 |
comment |
Describing a multitape Turing Machine that enumerates the set of $i$ such that $w_i$ is accepted by $M_i$ BY "describe" we mean give the operations that the T.M. should perform (in terms of how the tape heads move) in order to perform this task. I do understand the simulation process. |
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May 2 |
asked | Describing a multitape Turing Machine that enumerates the set of $i$ such that $w_i$ is accepted by $M_i$ |
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Apr 19 |
comment |
Abelian group admitting a surjective homomorphism onto an infinite cyclic group I just realised my stupid mistake! Just one more thing. How can I show that any element of $G$ is expressible as a product of two elements, the first of which is from $\ker(f)$ and the second from $im(g)$? |
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Apr 19 |
accepted | Studying the action of $GL(V)$ on the vector space $V$ |
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Apr 19 |
comment |
Studying the action of $GL(V)$ on the vector space $V$ Thank you very much! It is all clear now! |
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Apr 19 |
asked | Studying the action of $GL(V)$ on the vector space $V$ |
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Apr 19 |
comment |
Abelian group admitting a surjective homomorphism onto an infinite cyclic group I have tried to show that $\ker(g)$ is trivial, but I am failing to do so. My problem is that the $x$ you chose above might have finite order so you would have $nx=0$ with $n \neq 0$ thus the kernel will not be trivial. I cannot find a method to by-pass this problem. |
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Apr 17 |
accepted | $G$ a group and $H, K\mathrel{\unlhd}G$. Assuming that $H \cap K = \{1_G\}$ and $G = \langle H, K \rangle$, prove that $G \cong H \times K$ |
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Apr 16 |
comment |
Abelian group admitting a surjective homomorphism onto an infinite cyclic group I am so very sorry for not understanding this immediately. Here is my point of confusion. I cannot see how $G = \ker(f) \times im(g) \implies G \cong \ker(f) \times \Bbb Z$, since $im(g) \subset G$ so you would get $G \cong \ker(f) \times im(g) \subset \ker(f) \times G$ |
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Apr 16 |
asked | $G$ a group and $H, K\mathrel{\unlhd}G$. Assuming that $H \cap K = \{1_G\}$ and $G = \langle H, K \rangle$, prove that $G \cong H \times K$ |
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Apr 16 |
asked | Abelian group admitting a surjective homomorphism onto an infinite cyclic group |
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Apr 12 |
comment |
Generalised eigenvalue is eigenvalue if it is in the field Thank you very much for your help, Branimir! |
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Apr 12 |
comment |
Generalised eigenvalue is eigenvalue if it is in the field I see my mistake! Thank you very much for pointing this out! |
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Apr 12 |
accepted | Generalised eigenvalue is eigenvalue if it is in the field |
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Apr 12 |
asked | Generalised eigenvalue is eigenvalue if it is in the field |
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Mar 26 |
revised |
Imposing the topology of open rays in $\Bbb R$ deleted 4 characters in body |
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Mar 25 |
revised |
Imposing the topology of open rays in $\Bbb R$ added 88 characters in body |
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Mar 24 |
comment |
Imposing the topology of open rays in $\Bbb R$ Dear Brian, I would like to ask your permission to integrate your suggestions and hints into a complete answer in my original post. May I do so? |
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Mar 16 |
revised |
Imposing the topology of open rays in $\Bbb R$ added 69 characters in body |
