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 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 Extension of a group homomorphism 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!