# Finding determinant of matrix lie group

Given $G=\left\{A\in M_2(\mathbb{R})\mid A^\top XA = X\right\}$ where $X = \pmatrix{3&1\\1&1}$ and a Lie algebra $\mathfrak g=\left\{Y\in M_2(\mathbb{R})\mid Y^\top X+XY = 0\right\}$, how would I show that every element in $G$ does not have $\det=-1$?

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The matrix $$A = \pmatrix{0&\sqrt{3}\\\frac{1}{\sqrt3}&0}$$ has $\det{A} = - 1$ and satisfies $A^T X A = X$ unless I miscalculated....
(The Group in question can be thought of as the orthogonal group relative to the scalar product induced by $X$, which has two components (in the sense of topology), like the usual $O(2)$. These are precisely the counterimages of $1, -1$ respectively, under $\det$)
Yes, @John. Something is wrong. May be they assumed earlier that $G$ is the connectivity component containing the identity element? That is a subgroup. Often Lie theory concentrates on that subgroup. For example, the Lie algebra of that subgroup is the same as the Lie algebra of the whole group for the obvious reason that all the paths via $1_G$ are contained in that component. –  Jyrki Lahtonen Dec 31 '11 at 16:38
Hi Jyrki. I think you may be correct. Im reading your reasoning and it seems to make sense. Maybe they are talking about the Lie Algebra of $G$? It further states in the question to show that "hence $G$ has only one component." Assuming we ignore the determinant part of the question, is it possible (how would I show?) to show that $G$ has only 1 component? i.e if there is more than 1 component, det(A)= 1 or -1. –  John Dec 31 '11 at 17:12
$G$, if defined as in your post, does have two components, they won't go away. If you are asked to show that only one component exists an additional condition has to be imposed. That said, your original question does not involve the Lie Algebra in any way. What is true (as a possible additional condition) is that the image of this Lie algebra under the exponential map at $E$ is a connected Lie sub group of $G$ (namely the one for which $\det = 1$). –  user20266 Dec 31 '11 at 19:52