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could any one help me how to solve :

prove that there exist solution for the equation $x^2=y$ in identity component of a lie group. I dont know how to start this one, what is the specia; about component of identity? well for $y=e$ we get $x=e$

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It is not clear where this question comes from, can you give more context? One could imagine trying to trace solutions by moving $y$ around, starting at $y=e$ (where by the way there are in general more solutions than just $x=e$), which is an approach that would require a connected group. But it is not at all clear that this is actually a very promising angle of attack. – Marc van Leeuwen Sep 24 '12 at 8:27
component of identity is an Normal Subgroup of the Lie Group, so can we just try to see in a group this kind of equation has solution or not? – miosaki Sep 24 '12 at 10:34
up vote 1 down vote accepted

The result is false: the matrix $$ y=\begin{pmatrix}-1&0\\0&-2\end{pmatrix} $$ lies in the identity component of $G=\mathbf{GL}(2,\mathbf R)$ (which is the subset of matrices with positive determinant), but the equation $x^2=y$ for $x\in G$ has no solution, as can be checked by writing the equations for the matrix coefficients explicitly and showing the absence of real solutions.

See also this question relating this equation to the image of the exponential map, and this other question.

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