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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
Apr
12
comment Generalised eigenvalue is eigenvalue if it is in the field
Thank you very much for your help, Branimir!
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!
Apr
12
accepted Generalised eigenvalue is eigenvalue if it is in the field
Apr
12
asked Generalised eigenvalue is eigenvalue if it is in the field
Mar
26
revised Imposing the topology of open rays in $\Bbb R$
deleted 4 characters in body
Mar
25
revised Imposing the topology of open rays in $\Bbb R$
added 88 characters in body
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?
Mar
16
revised Imposing the topology of open rays in $\Bbb R$
added 69 characters in body
Mar
15
comment Imposing the topology of open rays in $\Bbb R$
Thank you very much for your most helpful comments!