2 GROUP ACTIONS

Definition 2.0.1. Let \(X\) be a set and \(G\) a group. Then \(G\) acts on \(X\) if for each of \(g\in G\), there is a function; \(\,\, \alpha _g :\, X\longrightarrow X\,\) such that

(i)
\(\alpha _g \cdot \alpha _h = \alpha _{gh}\,\) for \(g\,,\, h\in G\).
(ii)
\(\alpha _e = 1_X\)

Example 2.0.2.

1.
\(S_3 = \{(1)\,, \, (12)\, , \, (13)\, ,\, (23)\, , \, (123)\, ,\, (132)\}\)
\(X = \{(1)\, , \, (123)\, , \, (132)\}\)

Check: If \(\, \forall g\, , \, h\in S_3\,\) and \(\, x\in X\)

(i)
\(g\big (h(x)\big ) = \big (gh\big )(x).\) \(\hspace {0.5cm} 12(23(13)) = (23) = (12)(23)(13)\)
(ii)
\((1)(x) = x\)
2.
Trivial Action. Given a group \(G\), there exists the trivial action of \(G\) on \(G\) as follows; \(\, g(x) = x\,\, \forall g\in G\, , \,\, x\in G\).
(i)
Let \(g\, , \, h\in G\), and \(\, x\in G\). \(\, \,\,\, g\big (h(x)\big ) = g(x) = x = g\, h(x)\)
(ii)
\(e(x) = x\)
3.
Regular Action: Every group act on itself by multiplication on the left. \(G\times G\, \longmapsto G\,\) defined by; \(\, \, (g\, , \,x) \longmapsto gx\,\, \forall g\in G\, , \, x\in G\).
(i)
Let \(g\, , \, h\in G\) and \(x\in G\). Then \(g\big (h(x)\big ) = g\big (hx\big ) = \big (gh\big )(x)\)
(ii)
\(e(x) = e\cdot x = x\)
4.
Conjugation on Elements
Define \(\ g:\, x\longrightarrow gxg^{-1}\,\) where \(g\in G\, ,\, x\in X\). The map as defined is an action of \(G\) on \(X\).
(i)
Let \(g\, ,\, h\in G\) and \(x\in X\). Then \begin {align*} g\big (h(x)\big ) & = g\big (hxh^{-1}\big ) = ghxh^{-1}g^{-1}\\ & = (gh)x(gh)^{-1}\\ & = \big (gh\big )(x) \end {align*}
(ii)
\(e(x) = e x e^{-1} = x\)
5.
Conjugation on Subgroups
Define \(g:\, H\longmapsto gHg^{-1}\), where \(g\in G\) and \(H\leq G\)
(i)
Let \(g_1\, , \, g_2 \in G\). Then \begin {align*} g_1\big (g_2(H)\big ) & = \big \{g_1\big (g_2(h)\big )\, \big |\, h\in H\big \} = \big \{g_1\big (g_2hg_2^{-1}\big )\, \big |\, h\in H\big \}\\ & = \big \{g_1g_2hg_2^{-1}g_1^{-1}\,\big |\, h\in H\big \}\\ & = \big \{\big (g_1g_2\big )h\big (g_1g_2\big )^{-1}\, \big |\, h\in H\big \}\\ & = \big (g_1g_2\big )H \end {align*}
(ii)
\(eH = \big \{ehe^{-1}\, \big |\, h\in H\big \} = \big \{h\, \big |\, h\in H\big \} = H\)

Exercise 2.0.3. Show that the symmetric group \(S_n\) and its subgroups acts on the set \(X = \{1\, , \, 2\, , \ , \cdots \cdots \, , \, n\}\) by permuting its elements.

(i)
Let \(\, \alpha \, ,\, \beta \in S_n\), where \(\alpha \,,\, \in S_n\) and let \(x\in X\). Then \(\alpha \big (\beta (x)\big ) = \alpha \big (\beta x\big ) = \alpha \beta (x) = \big (\alpha \beta \big ) x\).

Definition 2.0.4. If \(G\) acts on a set \(SX\) and \(x\in X\) then,

(i)
The orbit of \(x\) denoted by \(O(x)\) or \(Orb_G(x)\) is defined by \(\, O(x) = \{gx|\, g\in G\}\).
(ii)
The stabilizer of \(X\), denoted by \(Gx\) or \(Stab_G(x)\) is defined by \(\, Gx = \{g\in G|\, gx = x\}\).

Note. The action of \(G\) on \(X\) is said to be transitive if there is only one orbit i.e for any \(x\, , y\in X\, , \exists \, g\in G\ni gx = y\).

Note. The stabilizer of \(X\) in \(G\) are all those elements of \(G\) that leave \(x\) fixed.

Theorem 2.0.5. The stabilizer of \(X\) in \(G\) is a subgroup of \(G\).

Proof. Clearly, \(G_X \neq \emptyset \), because \(e\in G_X\) since \(ex = x\). Now let \(g\, , h\in G_X\), we have
\(gh(x) = g\big (h(x)\big ) = g(x) = x \implies gh \in G_X^{-*}\). Next we show that \(g^{-1}\in G_X\) if \(g\in G_X\). \(\, g^{-1}(x) = g^{-1}\big (g(x)\big ) = \big (g^{-1}g\big )(x) = ex = x\implies g^{-1}\in G_X\). Now \(\, g^{-1}h(x) \in G_X\).

Example 2.0.6. The group \(G= \{(1)\, , \, (12)\,, \, (34)\, ,\, (12)(34)\}\) acts on the set \(X = \{1\, ,\, 2\, ,\, 3\, , \, 4\}\) \begin {align*} O(1) & = \{1\,,\, 2\} \hspace {0.4cm}\implies \, O(1) = (2)\\ O(2) & = \{1\, , \,2\}\\ O(3) & = \{3\, , \, 4\}\\ O(4) & = \{3\, , \, 4\} \end {align*}

\begin {align*} \text {Stabilizers are}\hspace {2cm} G_1 & = \{(1)\, , \, (34)\}\\ G_2 & = \{(1)\, , \, (34)\}\\ G_3 & = \{(1)\, , \, (12)\}\\ G_4 & = \{(1)\, , \, (12)\}\\\\ \end {align*}

Theorem 2.0.7 (Orbit Stabiliser Theorem). Let \(G\) be a finite group acting on a set \(X\) and let \(x\in X\), then \(\begin {vmatrix} O(x)\\ \end {vmatrix} = \big [G:\, G_X\big ]\,\) i.e \(\, \begin {vmatrix} G\\ \end {vmatrix} = \begin {vmatrix} O(x)\\ \end {vmatrix}\,\begin {vmatrix} G_x\\ \end {vmatrix}\).

Proof. Let \(G\big /G_X\) denote the family of cosets of \(G_X\) in \(G\). We want to exhibit a bijection \(\phi \) between \(O(x)\) and \(G\big /G_X\). Let \(y\in O(x)\), then \(y = g(x)\) for some \(g\in G\). Define \(\phi (y) = g G_X\). Clearly, \(\phi \) is single valued, because if \(y = h(x)\) for some \(h\in G\), then \(h^{-1}g(x) = x\) and \(h^{-1} g\in G_X\), hence \(h G_X = g G_X\). To see that \(\phi \) is one-one. Suppose \(\phi (y) = \phi (z)\), then \(\exists \,\) elements \(h\, , g\in G\ni y = g(x)\) and \(z = hx\) and \(g G_X = h G_X\). That is \(h^{-1}g\in G_X\). It follows that \(h^{-1} g x = x\) and so \(y = g(x) = h(x) = z\).
Finally, we show that \(\phi \) is onto. If \(g G_X\in G\big /G_X\), then let \(y = g x \in O(x)\) and note that \(\phi (y) = g G_X\). Thus we have the bijection \(\, \phi : \, O(x) \longrightarrow G\big /G_X \, \implies \, \begin {vmatrix} O(x)\\ \end {vmatrix} = \begin {vmatrix} G\big / G_X\\ \end {vmatrix}\). Thus the proof.

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