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Abstract Algebra Theory and Applications

Thomas W. Judson

Section 14.1 Groups Acting on Sets

Let \(X\) be a set and \(G\) be a group. A (left) action of \(G\) on \(X\) is a map \(G \times X \rightarrow X\) given by \((g,x) \mapsto gx\text{,}\) where
  1. \(ex = x\) for all \(x \in X\text{;}\)
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  2. \((g_1 g_2)x = g_1(g_2 x)\) for all \(x \in X\) and all \(g_1, g_2 \in G\text{.}\)
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Under these considerations \(X\) is called a \(G\)-set. Notice that we are not requiring \(X\) to be related to \(G\) in any way. It is true that every group \(G\) acts on every set \(X\) by the trivial action \((g,x) \mapsto x\text{;}\) however, group actions are more interesting if the set \(X\) is somehow related to the group \(G\text{.}\)
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Example 14.1.

Let \(G = GL_2( {\mathbb R} )\) and \(X = {\mathbb R}^2\text{.}\) Then \(G\) acts on \(X\) by left multiplication. If \(v \in {\mathbb R}^2\) and \(I\) is the identity matrix, then \(Iv = v\text{.}\) If \(A\) and \(B\) are \(2 \times 2\) invertible matrices, then \((AB)v = A(Bv)\) since matrix multiplication is associative.
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Example 14.2.

Let \(G = D_4\) be the symmetry group of a square. If \(X = \{ 1, 2, 3, 4 \}\) is the set of vertices of the square, then we can consider \(D_4\) to consist of the following permutations:
\begin{equation*} \{ (1), (1 \, 3), (2 \, 4), (1 \, 4 \, 3 \, 2), (1 \, 2 \, 3 \, 4), (1 \, 2)(3 \, 4), (1 \, 4)(2 \, 3), (1 \, 3)(2 \, 4) \}\text{.} \end{equation*}
The elements of \(D_4\) act on \(X\) as functions. The permutation \((1 \, 3)(2 \, 4)\) acts on vertex \(1\) by sending it to vertex \(3\text{,}\) on vertex \(2\) by sending it to vertex \(4\text{,}\) and so on. It is easy to see that the axioms of a group action are satisfied.
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In general, if \(X\) is any set and \(G\) is a subgroup of \(S_X\text{,}\) the group of all permutations acting on \(X\text{,}\) then \(X\) is a \(G\)-set under the group action
\begin{equation*} (\sigma, x) \mapsto \sigma(x) \end{equation*}
for \(\sigma \in G\) and \(x \in X\text{.}\)
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Example 14.3.

If we let \(X = G\text{,}\) then every group \(G\) acts on itself by the left regular representation; that is, \((g,x) \mapsto \lambda_g(x) = gx\text{,}\) where \(\lambda_g\) is left multiplication:
\begin{gather*} e \cdot x = \lambda_e x = ex = x\\ (gh) \cdot x = \lambda_{gh}x = \lambda_g \lambda_h x = \lambda_g(hx) = g \cdot ( h \cdot x)\text{.} \end{gather*}
If \(H\) is a subgroup of \(G\text{,}\) then \(G\) is an \(H\)-set under left multiplication by elements of \(H\text{.}\)
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Example 14.4.

Let \(G\) be a group and suppose that \(X=G\text{.}\) If \(H\) is a subgroup of \(G\text{,}\) then \(G\) is an \(H\)-set under conjugation; that is, we can define an action of \(H\) on \(G\text{,}\)
\begin{equation*} H \times G \rightarrow G\text{,} \end{equation*}
via
\begin{equation*} (h,g) \mapsto hgh^{-1} \end{equation*}
for \(h \in H\) and \(g \in G\text{.}\) Clearly, the first axiom for a group action holds. Observing that
\begin{align*} (h_1 h_2, g) & = h_1 h_2 g (h_1 h_2 )^{-1}\\ & = h_1( h_2 g h_2^{-1}) h_1^{-1}\\ & = (h_1, (h_2, g) )\text{,} \end{align*}
we see that the second condition is also satisfied.
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Example 14.5.

Let \(H\) be a subgroup of \(G\) and \({\mathcal L}_H\) the set of left cosets of \(H\text{.}\) The set \({\mathcal L}_H\) is a \(G\)-set under the action
\begin{equation*} (g, xH) \mapsto gxH\text{.} \end{equation*}
Again, it is easy to see that the first axiom is true. Since \((g g')xH = g( g'x H)\text{,}\) the second axiom is also true.
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If \(G\) acts on a set \(X\) and \(x, y \in X\text{,}\) then \(x\) is said to be \(G\)-equivalent to \(y\) if there exists a \(g \in G\) such that \(gx =y\text{.}\) We write \(x \sim_G y\) or \(x \sim y\) if two elements are \(G\)-equivalent.
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Proof.

The relation \(\sim\) is reflexive since \(ex = x\text{.}\) Suppose that \(x \sim y\) for \(x, y \in X\text{.}\) Then there exists a \(g\) such that \(gx = y\text{.}\) In this case \(g^{-1}y=x\text{;}\) hence, \(y \sim x\text{.}\) To show that the relation is transitive, suppose that \(x \sim y\) and \(y \sim z\text{.}\) Then there must exist group elements \(g\) and \(h\) such that \(gx = y\) and \(hy= z\text{.}\) So \(z = hy = (hg)x\text{,}\) and \(x\) is equivalent to \(z\text{.}\)
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If \(X\) is a \(G\)-set, then each partition of \(X\) associated with \(G\)-equivalence is called an orbit of \(X\) under \(G\text{.}\) We will denote the orbit that contains an element \(x\) of \(X\) by \({\mathcal O}_x\text{.}\)
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Example 14.7.

Let \(G\) be the permutation group defined by
\begin{equation*} G =\{(1), (1 \, 2 \, 3), (1 \, 3 \, 2), (4 \, 5), (1 \, 2 \, 3)(4 \, 5), (1 \, 3 \, 2)(4 \, 5) \} \end{equation*}
and \(X = \{ 1, 2, 3, 4, 5\}\text{.}\) Then \(X\) is a \(G\)-set. The orbits are \({\mathcal O}_1 = {\mathcal O}_2 = {\mathcal O}_3 =\{1, 2, 3\}\) and \({\mathcal O}_4 = {\mathcal O}_5 = \{4, 5\}\text{.}\)
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Now suppose that \(G\) is a group acting on a set \(X\) and let \(g\) be an element of \(G\text{.}\) The fixed point set of \(g\) in \(X\text{,}\) denoted by \(X_g\text{,}\) is the set of all \(x \in X\) such that \(gx = x\text{.}\) We can also study the group elements \(g\) that fix a given \(x \in X\text{.}\) This set is more than a subset of \(G\text{,}\) it is a subgroup. This subgroup is called the stabilizer subgroup or isotropy subgroup of \(x\text{.}\) We will denote the stabilizer subgroup of \(x\) by \(G_x\text{.}\)
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Remark 14.8.

It is important to remember that \(X_g \subset X\) and \(G_x \subset G\text{.}\)
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Example 14.9.

Let \(X = \{1, 2, 3, 4, 5, 6\}\) and suppose that \(G\) is the permutation group given by the permutations
\begin{equation*} \{ (1), (1 \, 2)(3 \, 4 \, 5 \, 6), (3 \, 5)(4 \, 6), (1 \, 2)( 3 \, 6 \, 5 \, 4) \}\text{.} \end{equation*}
Then the fixed point sets of \(X\) under the action of \(G\) are
\begin{gather*} X_{(1)} = X,\\ X_{(3 \, 5)(4 \, 6)} = \{1,2\},\\ X_{(1 \, 2)(3 \, 4 \, 5 \, 6)} = X_{(1 \, 2)(3 \, 6 \,5 \, 4)} = \emptyset\text{,} \end{gather*}
and the stabilizer subgroups are
\begin{gather*} G_1 = G_2 = \{(1), (3 \, 5)(4 \, 6) \},\\ G_3 = G_4 = G_5 = G_6 = \{(1)\}\text{.} \end{gather*}
It is easily seen that \(G_x\) is a subgroup of \(G\) for each \(x \in X\text{.}\)
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Proof.

Clearly, \(e \in G_x\) since the identity fixes every element in the set \(X\text{.}\) Let \(g, h \in G_x\text{.}\) Then \(gx = x\) and \(hx = x\text{.}\) So \((gh)x = g(hx) = gx = x\text{;}\) hence, the product of two elements in \(G_x\) is also in \(G_x\text{.}\) Finally, if \(g \in G_x\text{,}\) then \(x = ex = (g^{-1}g)x = (g^{-1})gx = g^{-1} x\text{.}\) So \(g^{-1}\) is in \(G_x\text{.}\)
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We will denote the number of elements in the fixed point set of an element \(g \in G\) by \(|X_g|\) and denote the number of elements in the orbit of \(x \in X\) by \(|{\mathcal O}_x|\text{.}\) The next theorem demonstrates the relationship between orbits of an element \(x \in X\) and the left cosets of \(G_x\) in \(G\text{.}\)
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Proof.

We know that \(|G|/|G_x|\) is the number of left cosets of \(G_x\) in \(G\) by Lagrange’s Theorem (Theorem 6.10). We will define a bijective map \(\phi\) between the orbit \({\mathcal O}_x\) of \(X\) and the set of left cosets \({\mathcal L}_{G_x}\) of \(G_x\) in \(G\text{.}\) Let \(y \in {\mathcal O}_x\text{.}\) Then there exists a \(g\) in \(G\) such that \(g x = y\text{.}\) Define \(\phi\) by \(\phi( y ) = g G_x\text{.}\) To show that \(\phi\) is one-to-one, assume that \(\phi(y_1) = \phi(y_2)\text{.}\) Then
\begin{equation*} \phi(y_1) = g_1 G_x = g_2 G_x = \phi(y_2)\text{,} \end{equation*}
where \(g_1 x = y_1\) and \(g_2 x = y_2\text{.}\) Since \(g_1 G_x = g_2 G_x\text{,}\) there exists a \(g \in G_x\) such that \(g_2 = g_1 g\text{,}\)
\begin{equation*} y_2 = g_2 x = g_1 g x = g_1 x = y_1; \end{equation*}
consequently, the map \(\phi\) is one-to-one. Finally, we must show that the map \(\phi\) is onto. Let \(g G_x\) be a left coset. If \(g x = y\text{,}\) then \(\phi(y) = g G_x\text{.}\)
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