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seen Mar 25 at 3:14

Jul
2
awarded  Curious
Mar
19
comment How to get $Y^2=Y$ but $YY^{-1}=I$?
Yes, to some extent, my question is gibberish. You can define what means $Y^2$ and $Y^{-1}$. Anyway, I want two DISTINCT objects, $Y$ and $I$, such that $Y^2=Y$ but $Y^{-1}=I$.
Mar
19
comment How to get $Y^2=Y$ but $YY^{-1}=I$?
This is a smart constructions. But why $YY^{-1}=\mathbb R$?
Mar
19
comment How to get $Y^2=Y$ but $YY^{-1}=I$?
I have not given the definitions on the operation $Y^2$ and $Y^{-1}$. I left these for free discussion. Of course, $Y=I$ is not the solution I want. I hope $Y$ and $I$ should be two distinct objects, either elements or subsets.
Mar
19
comment How to get $Y^2=Y$ but $YY^{-1}=I$?
You can define it by your self. At present, I define $Y^{-1}=\{y^{-1}:y\in Y\}$. But I have not found what I want.
Mar
19
asked How to get $Y^2=Y$ but $YY^{-1}=I$?
Mar
13
comment How to get the following reductions?
Wonderful! Thank you very much! I understand now.
Mar
12
comment How to get the following reductions?
Thanks a lot. But I have still failed to work out the second formula. Would you please to give me a detailed reduction?
Mar
12
asked How to get the following reductions?
Mar
3
comment Is it possible to make a commutative homomorphism image non-commutative?
Thanks. This comment is useful. But other solutions are still expected.
Mar
3
comment Is it possible to make a commutative homomorphism image non-commutative?
Thanks! But the start point is $G\neq K$ since $G$ is non-Abelian while $K$ is Abelian.
Mar
1
revised Is it possible to make a commutative homomorphism image non-commutative?
improved formatting
Mar
1
asked Is it possible to make a commutative homomorphism image non-commutative?
Mar
1
comment Extensions of homomorphisms
I encounter a similar but seemingly different question: Given a homomorphism $\phi: G\to K$ where $K$ is Abelian, how to get a new homomorphism $\varphi:\bar{G}\to\bar{K}$, such that: (1) $G$ and $K$ can be embedded into $\bar{G}$ and $\bar{K}$, respectively; (2) $\varphi(bar{G})$ is a non-Abelian subgroup of $\bar{K}$.
Nov
6
answered Which is easier to work out: determinant or inverse?
Sep
23
revised The probability of A and B getting to know each other
improved explanations
Sep
23
answered The probability of A and B getting to know each other
Sep
23
revised The probability of A and B getting to know each other
more precise tags
Sep
22
comment The probability of A and B getting to know each other
Do you mean the probability is near to zero?
Sep
22
comment The probability of A and B getting to know each other
Would you like to give me a detailed reduction process? I encounter this problem during my an ongoing manuscript. Thank you very much!