Skip to main content

Section 5.2 Self Actions and Conjugacy

Subsection Self Actions

“Selfless action is a source of strength.”
―Mahatma Ghandi

Example 5.11. Groups Acting on Themselves.

  • Left Regular Action.
    Let \(G\) be any group and define an action \(\cdot\) of \(G\) on itself by the rule
    \begin{equation*} g \cdot x = g x, \;\text{for}\; g \in G \;\text{and}\; x \in G. \end{equation*}
    This is also known as the left regular action
  • Self Conjugation Action.
    Let \(G\) be any group and fix an element \(g \in G\text{.}\) Define the conjugation action of \(G\) on itself by setting
    \begin{equation*} g\cdot x=gxg^{-1} \;\text{for any}\; g,x\in G. \end{equation*}
    This is also known as self-conjugation

Proof.

If \(\rho:G\to \operatorname{Perm}(G)\) is the permutation representation for \(G\) acting on \(G\) by Conjugation, then
\begin{equation*} g\in \operatorname{Ker}\rho\iff g\cdot x=x, \forall x\in G \iff gxg^{-1}=x, \forall x\in G \end{equation*}
\begin{equation*} \iff gx=xg, \forall x\in G \iff g\in Z(G). \end{equation*}
A group may also act on its cosets.

Example 5.14. Actions on Cosets.

For a subgroup \(H \leq G\text{,}\) consider the set of left cosets \(G/H = \{gH : g \in G\}\text{.}\) Then \(G\) acts on \(G/H\) my left multiplication.

Qual Watch.

Proving Lemma 5.15 is extremely similar to Part (a) of [cross-reference to target(s) "june-2016-1" missing or not unique] on the [cross-reference to target(s) "june-2016" missing or not unique] qualifying exam.

Proof.

Let \(S=G/H\) and note that \(|S|=p\text{.}\) Let \(K\) denote the kernel of the permutation representation generated by \(G\) acting on \(S\) by left multiplication.
The First Isomorphism Theorem tells us that \(G/K\cong\im(\rho)\leq S_p\text{.}\) Thus \(|K|\big|p!\) by Lagrange’s Theorem. Let \(k\in K\text{.}\) Then \(kgH=gH\) for all \(g\in G\text{,}\) making \(k\in gHg\inv\) for all \(g\in G,\) including \(e\text{.}\) Thus \(K\nsg H\text{.}\)
This yields \([G:K]=[G:H][H:K]\text{.}\) Let \(h=[H:K]\text{,}\) giving us \([G:K]=ph\text{.}\) As \(|K|\big|p!\) we have \(h|(p-1)!\text{,}\) so \(h<p\text{.}\) But \(p\) is the smallest prime dividing the order of \(G\text{,}\) and thus \(h=1\text{,}\) making \(H=K\) and \(H\nsg G\text{.}\)

Qual Watch.

Proving Theorem 5.16 was [cross-reference to target(s) "jan-2023-1" missing or not unique] on the [cross-reference to target(s) "jan-2023" missing or not unique] qualifying exam and [cross-reference to target(s) "jan-2015-1" missing or not unique] on the [cross-reference to target(s) "jan-2015" missing or not unique] qualifying exam.

Subsection Conjugacy

“Words may inspire but only action creates change.”
―Simon Sinek

Definition 5.17. Conjugate.

Let \(G\) be a group. Two elements \(g,g' \in G\) are conjugate if there is \(h \in G\) with \(g' = hgh^{-1}\text{.}\) Two subsets \(S,S' \subseteq G\) are conjugate if there is \(h \in G\) with \(S' = hSh^{-1}\text{.}\)

Definition 5.18. Conjugacy Class.

The conjugacy class of an element \(x \in G\) is \(C_G(x) = \{gxg^{-1} \mid g \in G\}\text{.}\)
We will not prove this result here, as it is really a specific case of Theorem 5.26, which we examine in the next section.

Example 5.20.

The conjugacy classes for \(S_4\) are
  1. \(\{e\}\text{,}\)
  2. all two cycles of which there are \({4 \choose 2} = 6\text{,}\)
  3. all three cycles of which there are \(4 \cdot 2 = 8\text{,}\)
  4. all four cycles of which there are \(3! = 6\text{,}\) and
  5. all product of two disjoint two cycles of which there are \(3\text{.}\)
Here is a convenient way of checking the conjugacy of two permutations.

Proof.

If two elements of \(S_n\) are conjugate, say \(\beta = \sigma\alpha \sigma^{-1}\text{,}\) then they have the same cycle type, since we may write \(\alpha\) as a product of disjoint cycles \(\alpha = \alpha_1 \cdots \alpha_m\) and then apply Lemma 2.7. Indeed, \(\sigma\alpha \sigma^{-1} = (\sigma\alpha_1 \sigma^{-1}) \cdots (\sigma\alpha_m \sigma^{-1})\) and the Lemma 2.7 shows that the right-side is a product of disjoint cycles.

Proof.

Define the conjugation action of \(G\) on \(N\) by \(g\cdot n=gng^{-1}\) for all \(g\in G\) and \(n\in N\text{.}\) Since \(N\nsg G\) this is well defined. The two properties in the definition of the action hold for the action of \(G\) by Conjugation on \(N\) since they hold more generally for the action of \(G\) by conjugation on \(G\text{.}\) Therefore this is indeed an action. The orbits of elements \(n\in N\) under this action are \(C_G(n)\text{.}\) Thus the conjugacy classes of the elements of \(N\) partition \(N\text{.}\)