# Group Action on a Set

#### Oxymoron

Question

Let $G=GL(2,\mathbb{R})$ be the group of invertible $2\times 2$ martrices with real entries. Consider the action of $G$ on itself by conjugation. For the element

$$A= \left(\begin{array}{cc} 2 & 1 \\ 0 & 3 \end{array}\right)$$

of $G$, describe (i) the orbit and (ii) the isotropy group of $A$

Sorry, I have no working out because I am completely stumped. Can anyone give me some helpful hints or pointers. Thanks

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#### Oxymoron

The orbit is going to be of the form

$$O_A = \{gA\,|\, g\in GL(2,\mathbb{R})\}$$

Here $g$ can be found by the following way

$$gA = Ag$$ where $$g \in GL(2,\mathbb{R})$$

$$\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)\left(\begin{array}{cc}2 & 1 \\ 0 & 3\end{array}\right) = \left(\begin{array}{cc}2 & 1 \\ 0 & 3\end{array}\right)\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)$$

$$\left(\begin{array}{cc}2a & a+3b \\ 2c & c+3d\end{array}\right) = \left(\begin{array}{cc}2a+c & 2b+d \\ 3c & 3d\end{array}\right)$$

Which implies that $2c = 3c = 0 \Leftrightarrow c = 0$. Hence

$$\left(\begin{array}{cc}2a & a+3b \\ 0 & 3d\end{array}\right) = \left(\begin{array}{cc}2a & 2b+d \\ 0 & 3d\end{array}\right)$$

And we can write

$$g = \left(\begin{array}{cc}a & d-a \\ 0 & d\end{array}\right)$$

Therefore the orbit can be described as

$$O_A = \{gA\,|\, g\in GL(2,\mathbb{R})\}$$

So

$$O_A = \left(\begin{array}{cc}2a & -2a + 3d \\ 0 & 3d\end{array}\right)\quad \forall a,b \in \mathbb{R}$$

how does this look?

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#### Oxymoron

The isotropy group is the subgroup of $GL(2,\mathbb{R})$ consisting of the elements that do not move $A$. That is

$$G_A = \{gA = A\,|\,g\in GL(2,\mathbb{R})\}$$

Therefore we have

$$\left(\begin{array}{cc}2a & -2a+3d \\ 0 & 3d\end{array}\right) = \left(\begin{array}{cc}2 & 1 \\ 0 & 3\end{array}\right)$$

So $a = 1, \, d = 1$. Therefore

$$G_A = \left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right) = e$$

the isotropy subgroup consists of the identity element.

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#### Oxymoron

Anyone know if I have done this correctly? Anyone?

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