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Is it possible to find a linear mapping $l$ (from set $L$ to set $L'$), where $l(x^{-1})\neq(l(x))^{-1}$, $x\in L$ ?

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Do you want a hint or an answer? If just a hint, then $l$ is linear means $l(0)=0$. – Todd Wilcox Nov 21 '12 at 15:41
up vote 2 down vote accepted

No, Linear mapping must hold $l(x y) = l(x) l(y)$ and also $l(1)=1'$ Thus you get $l(1)=l(x x^{-1})=l(x)l(x^{-1})=1'\Rightarrow l(x^{-1})= l(x)^{-1}$

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To clarify the use of $0$ in my hint and $1$ in 44874's answer, in both cases we are talking about the identity elements for the groups $L$ and $L'$. If the group operation is seen as "additive", then $0$ is conventionally used, and likewise for "multiplicative" operations and $1$. Either way it amounts to two different notations for the same object. If you are studying undergraduate or high school linear algebra, you have probable seen $0$ used in this case. – Todd Wilcox Nov 21 '12 at 15:47

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