For each standard action below, identify the orbits, stabilizers, global fixed points, kernel, and image in the relevant symmetric group:
- left translation of \(G\) on itself;
- conjugation of \(G\) on itself;
- conjugation of \(G\) on its set of subgroups;
- left translation of \(G\) on \(G/H\) for a fixed subgroup \(H\le G\).
Solution
1. Left translation on \(G\).
The action is transitive, so there is one orbit, namely \(G\). The stabilizer of every \(x\in G\) is trivial because \[ gx=x\iff g=e. \] Hence the kernel is trivial and the image is the left regular copy of \(G\) in \(\Sym(G)\). If \(G\ne1\), there are no global fixed points.
2. Conjugation on \(G\).
The orbit of \(x\) is its conjugacy class \[ \operatorname{Cl}_G(x). \] Its stabilizer is the centralizer \[ C_G(x). \] The global fixed points are exactly \(Z(G)\). The kernel consists of elements conjugating every \(x\) trivially, hence is also \(Z(G)\). Therefore the image is \[ \Inn(G)\cong G/Z(G). \]
3. Conjugation on the set of subgroups.
The orbit of \(H\) is the conjugacy class of the subgroup, \[ \{gHg^{-1}:g\in G\}. \] The stabilizer is its normalizer \[ N_G(H). \] The global fixed points are precisely the normal subgroups of \(G\). The kernel is \[ \bigcap_{H\le G}N_G(H), \] the subgroup of elements normalizing every subgroup of \(G\). Thus the image is the quotient of \(G\) by this kernel.
4. Left translation on \(G/H\).
The action is transitive. The stabilizer of the coset \(xH\) is \[ xHx^{-1}. \] The kernel is the core of \(H\), \[ \operatorname{core}_G(H) =\bigcap_{x\in G}xHx^{-1}, \] the largest normal subgroup of \(G\) contained in \(H\). Hence the image is \[ G/\operatorname{core}_G(H). \] If \(H<G\), there is no global fixed coset; if \(H=G\), the unique coset is fixed.