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I am trying to get a grasp on what a representation is, and a professor gave me a simple example of representing the group $Z_{12}$ as the twelve roots of unity, or corresponding $2\times 2$ matrices. Now I am wondering how $\operatorname{GL}(1,\mathbb{C})$ and $\operatorname{GL}(2,\mathbb{R})$ are related, since the elements of both groups are automorphisms of the complex numbers. $\operatorname{GL}(\mathbb{C})$, the group of automorphisms of C, is (to my understanding) isomorphic to both $\operatorname{GL}(1,\mathbb{C})$ and $\operatorname{GL}(2,\mathbb{R})$ since the complex numbers are a two-dimensional vector space over $\mathbb{R}$. But it doesn't seem like these two groups are isomorphic to each other.

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    Okay. Sorry if that was rude or anything. I know very little about representation theory, and I definitely didn't know that people study representations of Lie groups totally independently of Lie algebras.2010-12-01

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No, they're not isomorphic. $GL(1,\mathbb{C})$ injects into $GL(2, \mathbb{R})$ but the latter consists of the set of real automorphisms of $\mathbb{R}^2$, while the former consists of the set of all complex automorphisms of $\mathbb{C}$. For instance, an element of $GL(1, \mathbb{C})$ is the composite of a scaling (dilation) by a positive real number and a rotation, so is $\mathbb{R}_{>0} \times SO(2) = \mathbb{R}_{>0} \times S^1$. However, $GL(2, \mathbb{R})$ includes more complicated things (e.g. reflections, things which even if scaled do not become orthogonal transformations).

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    @Pete: Fixed, tha$n$ks.2010-11-30
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The one group is commutative, the other is not. So they can not be isomorphic.

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$GL(1,\mathbb{C})$ can be thought of as the group of automorphisms of $\mathbb{C}$ as a complex vector space. Each such automorphism is given by multiplication by a nonzero complex number. On the other hand, $GL(2,\mathbb{R})$ can be thought of as the group of automorphisms of $\mathbb{C}$ as a real vector space, and hence it is larger. In terms of matrices, $GL(1,\mathbb{C})$ corresponds to all $2$-by-$2$ real matrices of the form $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ with $a^2+b^2\gt0$, which is a proper subgroup of the group $GL(2,\mathbb{R})$ of all $2$-by-$2$ real invertible matrices.

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    @Sam: the same mathematical object often has several different automorphism groups, depending upon what kind of automorphisms are being considered. For instance the automorphism group of $\mathbb{R}$ as a metric space (i.e., its isometry group) is much smaller than its automorphism group as a topological space, which is (very!) much smaller than its automorphism group as a set (i.e., the bijections). Something similar is happening here: a complex vector space has "more structure" than its underlying real vector space, hence fewer automorphisms.2011-06-07
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I'd like to supplement the above answers with two important examples. Consider a hyperbolic transformation $H(x, y) = (2x, y/2)$ and a shear $S(x,y) = (x+y, y)$, both elements of $GL(2,R)$. If you draw how these act on points of $R^2$ you will see that these distort angles. On the other hand, elements of $GL(1,C)$ always preserve angles (and thus so do holomorphic maps, away from critical points).

That is, the extra algebraic structure preserved by $GL(1,C)$ translates to the geometric property of preserving angles and "handedness".