📂Abstract Algebra

Isotropic Subgroups

Definition ¹

Let’s consider $G$ as a group, and define $X$ as a $G$-set. For $x \in X$ and $g \in G$, let $X_{g} := \left\{ x \in X \ | \ gx = x \right\}$ and $G_{x} := \left\{ g \in G \ | \ gx = x \right\}$. We define the Isotropy Subgroup of $G$ with respect to $x$ as $G_{x}$.

Explanation

To get a grasp of what an isotropy subgroup is, one needs to understand the action of a group.

For the set of points and lines on the left side of the above figure $$ X : = \left\{ 1,2,3,4 , C, p_{1}, p_{2} , p_{3} , p_{4} , s_{1}, s_{2} , s_{3} , s_{4} , d_{1}, d_{2} , m_{1} , m_{2} \right\} $$ consider the icosahedral group $G = D_{4}$. $X$ becomes a $D_{4}$-set. In simple terms, $X_{g}$ is a set unaffected by $g$, while $G_{x}$ becomes a subgroup that cannot influence $x$. Here, $C$ is the center of $X$, and $\rho_{0}$ is the identity of $G$, which for every $x,g$, satisfies $C \in X_{g}$ and $\rho_{0} \in G_{x}$.

Let’s verify whether the examples below actually align with the figure above: $$ \begin{align*} X_{\rho_{0}} =& X \\ X_{\rho_{1}} =& C \\ X_{\mu_{1}} =& \left\{ C, p_{1}, p_{3} , s_{1} , s_{3} , m_{1} , m_{2} \right\} \\ G_{1} =& \left\{ \rho_{0} , \delta_{2} \right\} \\ G_{s_{3}} =& \left\{ \rho_{0} , \mu_{1} \right\} \\ G_{d_{1} } =& \left\{ \rho_{0} , \rho_{2} , \delta_{1} , \delta_{2} \right\} \end{align*} $$ Here, when there exists $g \in G$ satisfying $g x_{1} = x_{2}$ for $x_{1} , x_{2} \in X$, let’s define the equivalence relation $x_{1} \sim x_{2}$. If $x$ belongs to the cell of the partition by $\sim$, that cell is called the orbit of $x$ and is denoted as $Gx$. Though it might look complicated, it simply means that $X$ is split into parts like $\left\{ C \right\} , \left\{ 1,2,3,4 \right\} , \left\{ p_{1}, p_{2} , p_{3} , p_{4} \right\} , \left\{ s_{1}, s_{2} , s_{3} , s_{4} \right\} , \left\{ d_{1}, d_{2} \right\} , \left\{ m_{1} , m_{2} \right\}$.

It’s important to note that $G_{x}$ and $Gx$ are distinguished by whether they are subscripts or not. For example, $G_{1}$ acts as an isotropy subgroup that cannot influence $1$, which is $G_{1} = \left\{ \rho_{0} , \delta_{2} \right\}$, and $G1$ is the orbit of $1$, which is $G1 = \left\{ 1,2,3,4 \right\}$.

Generally speaking, consider the following theorem. It looks like it reminds us of Lagrange’s theorem, but a closer look at the notation reveals it’s somewhat unrelated.

Theorem

If $X$ is a $G$-set, then $|Gx| = ( G : G_{x})$ holds. If $G$ is a finite group, then $|Gx|$ is a divisor of $|G|$.

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• $( G : G_{x})$ is the number of cosets of $G_{x}$, and $G / G_{x}$ is the set of cosets of $G_{x}$.

Proof

Assume $x_{1} \in Gx$, then there exists $g_{1} \in G$ satisfying $g_{1} x = x_{1}$. When $ \psi : Gx \to G / G_{x}$ is defined as $\psi (x_{1}) := g_{1} G_{x}$, showing that $\psi$ is a bijection completes the proof.

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Part 1. $\psi$ is a function.

Assuming $g_{1} x = x_{1}$, suppose there exists $g_{1} ' \in G_{x}$ satisfying $g_{1}’ x = x_{1}$. Then, $g_{1} x = g_{1}’ x$ and $x = (g_{1}^{-1} g_{1} ’ ) x$, thus $(g_{1}^{-1} g_{1} ’ ) \in G_{x}$ must hold. Therefore, $g_{1} ' \in g_{1} G_{x}$, and since $g_{1} G_{x} = g_{1}’ G_{x}$, $\psi$ is a function.

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Part 2. $\psi$ is injective.

If $x_{1} , x_{2} \in Gx$ and assuming $\psi (x_{1} ) =\psi (x_{2} ) $, then there exists $g_{1} , g_{2} \in G$ satisfying $x_{1} =g_{1} x$, $x_{2} = g_{2} x$, and $g_{2} \in g_{1} G_{x}$. For some $g \in G_{x}$, $g_{2} = g_{1} g$, and $x_{2} = g_{2} x = g_{1} (g x) = g_{1} x = x_{1}$ holds. Since $\psi (x_{1} ) =\psi (x_{2} ) \implies x_{1} = x_{2}$, $\psi$ is injective.

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Part 3. $\psi$ is surjective.

For a coset of $G_{x}$, $g_{1} G_{x}$, if $g_{1} x = x_{1}$, then $g_{1} G_{x} = \psi ( x_{1} )$ holds. Since $x_{1} \in Gx$ is valid for all $\psi ( x_{1} )$, $\psi$ is surjective.

Parts 1~3 demonstrated that $\psi$ is a bijection, thereby proving the following: $$ | Gx | = |G / G_{x}| = ( G : G_{x} ) $$

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Part 4.

If $|G| < \infty$, then $|G| = |G_{x} | (G : G_{x}) = |G_{x} | |Gx|$ holds, making $|Gx|$ a divisor of $|G|$.

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