3.13 Actions of \(G\) on itself
In this section, \(G\) and \(H\) are groups with identity elements written as \(1\).
Here are a couple of simple lemmas I probably should have stated earlier.
Suppose \(f:G\to H\) is a one-one homomorphism. Then the map \(G\to f(G)\) given by \(g\mapsto f(g)\) is an isomorphism of \(G\) onto its image \(f(G)\leq H\).
Exercise.
Suppose \(f:X\to Y\) is a bijection of sets. Then the map \(\sigma \mapsto f\sigma f^{-1}\) defines an isomorphism of groups \(A(X)\to A(Y)\).
Exercise.
There are various ways that a group can act on itself. The first interesting one is called the left action. It is used in the proof of the following theorem.
Let \(a:G\times G\to G\) be the map \(a(g,h) = gh\). Then \(a\) defines a group action, called the left action of \(G\) on itself, with associated homomorphism \(\lambda :G\to A(G)\) given by \(\lambda (g)(h) = gh\). The homomorphism \(\lambda \) is one-one. Consequently, \(G\) is isomorphic to the subgroup \(\lambda (G)\leq A(X)\).
In particular, every finite group of order \(n\) is isomorphic to a subgroup of the symmetric group \(S_n\)
To see that \(a\) is an action we have to check that the map \(G\times G\to G\) given by \((g,h)\mapsto gh\) satisfies Theorem 3.82 1 and 2. Both of these are obvious.
To check that \(\lambda \) is one-one, we need to check that \(\ker \lambda = 1\). Or in other words, we need to check that \(\lambda (g) = \operatorname{\mathrm{id}}_G\) iff \(g = 1\). So take \(g\in G\) and \(h\in H\). Then \(\lambda (g) = \operatorname{\mathrm{id}}_G \Rightarrow gh = \lambda (g)(h) = \operatorname{\mathrm{id}}_G(h) = h\Rightarrow g = 1\). So we get that \(\ker \lambda = 1\). Therefore, by Lemma 3.87, \(G\cong \lambda (G)\).
If \(|G| = n\), then we can find a bijection \(f:G\to [n]\), where \([n] = \{ 1,\ldots , n\} \). Then use Lemma 3.88
Recall from Definition 3.1 that a group automorphism of \(G\) is a homomorphism \(f:G\to G\) from \(G\) to itself, which is one-one and onto.
Write \(\operatorname{\mathrm{Aut}}G\) for the set of all group automorphisms of \(G\).
We have \(\operatorname{\mathrm{Aut}}G\leq A(G)\).
For \(g\in G\), let \(\iota (g): G\to G\) denote that map \(h\mapsto ghg^{-1}\). Then, for all \(g\in G\), \(\iota (g)\in \operatorname{\mathrm{Aut}}G\), and the resulting map \(\iota : G\to \operatorname{\mathrm{Aut}}G\) is a group homomorphism.
We have \(\iota (g)\in E(G)\) for all \(g\in G\). Let’s check that the map \(\iota :G\to E(G)\) is a monoid homomorphism so that we can use Lemma 3.78.
If \(g,h,k\in G\), then \(\iota (gh)(k) = gh(k)h^{-1}k^{-1} = g(hkh^{-1})g^{-1} = \iota (g)(\iota (h) k) = (\iota (g)\circ \iota (h))(k)\). Since this holds for all \(k\), \(\iota (gh) = \iota (g)\circ \iota (h)\). So \(\iota :G\to E(G)\) is a magma homomorphism. On the other hand, \(\iota (1)(h) = 1h1^{-1} = h\) for all \(h\in G\). So \(\iota (1) = \operatorname{\mathrm{id}}_G\), and, therefore, \(\iota :G\to E(G)\) is a monoid homomorphism.
Now, using Lemma 3.78, we get that \(\iota (G)\subseteq E(G)^{\times } = A(G)\), and that \(g\mapsto \iota (g)\) defines a group homorphism \(G\to A(G)\). Let’s show that the image lands in \(\operatorname{\mathrm{Aut}}(G)\).
Suppose \(g, h, k\in G\). Then \(\iota (g)(hk) = g(hk)g^{-1} = (ghg^{-1})(gkg^{-1}) = (\iota (g)(h))(\iota (g)(k))\). So, \(\iota (g):G\to G\) is a group homomorphism. Moreover, \(\iota (g)\) is a bijection with inverse \(\iota (g^{-1})\). So, in fact, \(\iota (g)\in \operatorname{\mathrm{Aut}}G\).
A group automorphism of \(G\) of the form \(\iota (g)\) is called an inner automorphism of \(G\). We write \(\operatorname{\mathrm{Inn}}G = \{ \rho (g):g\in G\} \subseteq \operatorname{\mathrm{Aut}}G\) for the set of inner automorphisms.
We have \(\operatorname{\mathrm{Inn}}G = \iota (G)\). Consequently, \(\operatorname{\mathrm{Inn}}G\leq \operatorname{\mathrm{Aut}}G\).
The kernel of the map \(\iota :G\to \operatorname{\mathrm{Aut}}G\) is the center, \(Z(G)\), of \(G\). Consequently, \(G/Z(G)\cong \operatorname{\mathrm{Inn}}G\).
Suppose \(g\in \ker \iota \). Then, for all \(h\in G\), we have \(gh = ghg^{-1}g = (\iota (g)(h))g = (\operatorname{\mathrm{id}}_G(h))g = h g\). So, by definition, \(g\in Z(G)\). On the other hand, if \(g\in Z(G)\), then, for all \(h\in G\), \(ghg^{-1} = hgg^{-1} = h\).
Since, \(\operatorname{\mathrm{Inn}}G = \iota (G)\), we get that \(\iota (G) \cong G/Z(G)\).