17 Group Actions

#todo/school : Problems 4, 5, 6, 17 📅 2025-01-22 ✅ 2025-01-21

4

Question

Let $G$ be a Group acting on a set $A$ and fix some $a \in A$ . Show that the following sets are Subgroups of $G$ :
a. The kernel of the action.
b. ${g \in G | g a = a}$ - this subgroup is called the stabilizer of $a$ in $G$ .

Proof

Let $σ : G \times A \to A$ be this group action.

a. Notice $Ker (σ) = {g \in G | g a = a \forall a \in A}$ . Notice that it's a subset of $G$ , a group already. Thus we just have to show closure under the binary operation, and under the inverse.

For the former, let $g_{1}, g_{2} \in Ker (σ)$ . Then consider the element $g_{1} g_{2}$ acting on some $a \in A$ . Namely, notice that:

\begin{aligned} (g_{1} g_{2}) \cdot a & = g_{1} \cdot (g_{2} \cdot a) \\ = g_{1} \cdot a & (g_{2} \in Ker (σ)) \\ = a & (g_{1} \in Ker (σ)) \end{aligned}

Thus then $g_{1} g_{2} \in Ker (σ)$ , showing closure under the binary operation.

Similarly, let $g \in Ker (σ)$ . We want to show $g^{- 1} \in Ker (σ)$ . Notice for any $a \in A$ :

\begin{aligned} g \cdot a & = a \\ g^{- 1} \cdot (g \cdot a) & = g^{- 1} \cdot a \\ (g^{- 1} g) \cdot a & = g^{- 1} \cdot a \\ 1_{G} \cdot a & = g^{- 1} \cdot a \\ a & = g^{- 1} \cdot a \end{aligned}

Thus $g^{- 1} \in Ker (σ)$ . Therefore, then $Ker (σ)$ is a subgroup of $G$ .

b. Consider $C = {g \in G | g a = a}$ . Notice that this is very similar to $Ker (σ)$ , except it is for our specific fixed $a \in A$ . The work to show it is a subgroup would be a copy of what we did for (a), so then WLOG then $C$ is a subgroup of $G$ .

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5

Question

Prove that the kernel of an action of the Group $G$ on the set $A$ is the same as the kernel of the corresponding permutation representation $G \to S_{A}$ . (see also 16 Homomorphisms, Isomorphisms#14)

Proof

Say $σ : G \times A \to A$ is such an action of the group $G$ on set $A$ . Then let $K = Ker (σ) = {g \in G | g \cdot a = a \forall a \in A}$ . Let $φ : G \to S_{A}$ be a group Homomorphism (or Group Morphism) from $G$ to $S_{A}$ .

We want to show that $K = Ker (φ)$ . Note here that $Ker (φ) = {g \in G | φ (g) = (1)}$ :

$(\subseteq)$ : Let $k \in K$ . Then $\forall a \in A$ we get that $k \cdot a = a$ . Notice then that letting $a \in A$ :

φ (k) \cdot a = σ_{k} (a) = k \cdot a = a

Thus $φ (k) = (1)$ so then $k \in Ker (φ)$ .

$(\supseteq)$ : Let $k \in Ker (φ)$ , so then $φ (k) = (1)$ . Let $a \in A$ . Then notice that:

φ (k) \cdot a = k \cdot a = a

since $φ (k)$ is the identity, and $φ (k) = σ_{k} = k$ . Therefore, then $k \in K$ by definition.

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6

Question

Prove that a Group $G$ acts faithfully (ie: the action's associated permutation representation is injective) (see Faithful Action) iff the kernel of the action is the set consisting only of the identity.

Proof

$(\Rightarrow)$ : Suppose $G$ acts faithfully. Namely, let $X$ be the set it acts faithfully on. Thus, then any $g \in G$ gives an injective permutation representation $φ : G \to S_{X}$ .

Consider the kernel of the action $Ker (σ) = {g \in G | g \cdot x = x \forall x \in X}$ . Clearly the identity $1_{G}$ is in this kernel since $1_{G} \cdot x = x$ by definition of a group action.

Now, assume for contradiction that there is some $g \neq 1_{G}$ where $g \in Ker (σ)$ . Thus $g \cdot x = x$ for any $x \in X$ . Because $1_{G} \cdot x = x = g \cdot x$ then $σ_{1_{G}} (x) = σ_{g} (x)$ . $x$ here is arbitrary, so then $φ (1_{G}) = φ (g)$ . But $φ$ is an injective map (by our supposition)! Thus then $1_{G} = g$ which is a contradiction. Thus, then the only element in the kernel can be the identity element.

$(\Leftarrow)$ : Suppose that $G$ is not acting faithfully. Thus $\exists g_{1} \neq g_{2} \in G$ such that $φ (g_{1}) = φ (g_{2})$ . We want to show that the kernel is not trivial here. These are permutations which can be inverted so then $φ (g_{1})^{- 1} φ (g_{2}) = (1)$ . Since $φ$ is a homomorphism, then $(1) = φ (g_{1}^{- 1} g_{2})$ so then $g_{1}^{- 1} g_{2} \in Ker (σ)$ . Notice that this element cannot be the identity since if it was then $g_{1} = g_{2}$ as a contradiction. Thus the kernel is not trivial.

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17

Question

Let $G$ be a group and let $G$ act on itself by left conjugation, so each $g \in G$ maps $G \to G$ by:

x \mapsto g x g^{- 1}

For fixed $g \in G$ , prove that conjugation by $g$ is an Isomorphism from $G$ onto itself (see 16 Homomorphisms, Isomorphisms#20). Deduce that $x$ and $g x g^{- 1}$ have the same order for all $x \in G$ and that for any subset $A \subseteq G$ then $| A | = | g A g^{- 1} |$ where $g A g^{- 1} = {g a g^{- 1} | a \in A}$ .

Proof

Let $φ$ be the isomorphism we are showing, for our fixed $g \in G$ (so $φ (x) = g x g^{- 1}$ for any $x \in G$ ). For the first condition of being a Homomorphism (or Group Morphism), let $x_{1}, x_{2} \in G$ be arbitrary. Then:

\begin{aligned} φ (x_{1} x_{2}) & = g (x_{1} x_{2}) g^{- 1} \\ = (g x_{1}) (x_{2} g^{- 1}) \\ = (g x_{1}) (g^{- 1} g) (x_{2} g^{- 1}) \\ = (g x_{1} g^{- 1}) (g x_{2} g^{- 1}) \\ = φ (x_{1}) φ (x_{2}) \end{aligned}

For being bijective:

Injective: let $x_{1}, x_{2} \in G$ and suppose $φ (x_{1}) = φ (x_{2})$ . Then $g x_{1} g^{- 1} = g x_{2} g^{- 1}$ so using cancellation rules then $x_{1} = x_{2}$ .
Surjective: let $y \in G$ . Choose $x = g^{- 1} y g \in G$ so then $φ (x) = g (g^{- 1} y g) g^{- 1} = (g g^{- 1}) y (g g^{- 1}) = y$ .

To show that $| x | = | g x g^{- 1} |$ assume $| x | = n$ . Then $x^{n} = 1$ . Then:

(g x g^{- 1})^{n} = g x^{n} g^{- 1} = φ (x^{n}) = φ (1_{G}) = 1_{G}

Thus $| g x g^{- 1} | = n = | x |$ .

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