555 reputation
217
bio website chicago.academia.edu/…
location Chicago, IL
age 23
visits member for 1 year, 7 months
seen Jun 15 at 14:51

I have a B.A. in Mathematics from the University of Chicago. I am starting a Ph.D. in Linguistics on September 2014 in the same university.


Jul
2
awarded  Curious
Jan
11
awarded  Yearling
Nov
5
awarded  Notable Question
Oct
9
asked $\beta$ - conversion and $\alpha$-reduction problem in $\lambda$-calculus
Oct
6
comment Is this $\beta$-reduction well defined?
The question is whether it is possible to operate as it is.
Oct
6
asked Is this $\beta$-reduction well defined?
Sep
7
awarded  Popular Question
Jul
4
comment A step in the proof of the Riemann Mapping Theorem
Are we considering the graph of $sin(1/x)$ as a subset of $\overline{\Bbb C}$?
Jul
4
revised A step in the proof of the Riemann Mapping Theorem
added 273 characters in body
Jul
4
asked A step in the proof of the Riemann Mapping Theorem
May
6
awarded  Caucus
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 $im(g)$?
Apr
19
accepted Studying the action of $GL(V)$ on the vector space $V$
Apr
19
comment Studying the action of $GL(V)$ on the vector space $V$
Thank you very much! It is all clear now!
Apr
19
asked Studying the action of $GL(V)$ on the vector space $V$
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.
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 $G \cong H \times K$
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 $G \cong \ker(f) \times im(g) \subset \ker(f) \times G$
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 $G \cong H \times K$
Apr
16
asked Abelian group admitting a surjective homomorphism onto an infinite cyclic group