Let G be a finite group, and for each positive integer n, let
\[ X _ { n } = \{ ( g _ { 1 } , . . . , g _ { n } ) : g _ { i } g _ { j } = g _ { j } g _ { i } \forall i , j \} . \]
Show that the formula
\[ h \cdot ( g _ { 1 } , \ldots , g _ { n } ) = ( h g _ { 1 } h ^ { - 1 } , \ldots , h g _ { n } h ^ { - 1 } ) , \]
defines an action of \(G\) on \(X _ { n }\) , and that \(| X _ { n + 1 } | = | G | \cdot | X _ { n } / G |\) for all n, where \(X _ { n } / G\) denotes the set of G-orbits in \(X _ { n }\)
Solution
For \(h\in G\), write \[ X_n^h\coloneqq \{x\in X_n:h\cdot x=x\} \] for the fixed-point set of \(h\).
1. Simultaneous conjugation preserves \(X_n\).
Let \((g_1,\ldots,g_n)\in X_n\) and \(h\in G\). For every \(i,j\), \[ \begin{aligned} (hg_ih^{-1})(hg_jh^{-1}) &=h(g_ig_j)h^{-1}\\ &=h(g_jg_i)h^{-1}\\ &=(hg_jh^{-1})(hg_ih^{-1}). \end{aligned} \] Thus the conjugated tuple again has pairwise commuting coordinates and lies in \(X_n\).
2. The displayed formula defines an action of \(G\) on \(X_n\).
Step 1 shows that the formula is well-defined on \(X_n\). The identity element acts trivially. For \(h,k\in G\) and \((g_1,\ldots,g_n)\in X_n\), coordinatewise conjugation gives \[ h\cdot\bigl(k\cdot(g_1,\ldots,g_n)\bigr) = (hk)\cdot(g_1,\ldots,g_n). \] These are precisely the two action axioms.
3. For each \(h\in G\), the elements of \(X_n^h\) are exactly the pairwisecommuting \(n\)-tuples whose entries all commute with \(h\).
For \((g_1,\ldots,g_n)\in X_n\), \[ h\cdot(g_1,\ldots,g_n)=(g_1,\ldots,g_n) \] if and only if \[ hg_ih^{-1}=g_i \] for every \(i\), equivalently \(hg_i=g_ih\) for every \(i\).
4. One has\[ \abs{X_{n+1}} = \sum_{h\in G}\abs{X_n^h}. \]
Partition \(X_{n+1}\) according to its last coordinate \(h\). For fixed \(h\in G\), a tuple \[ (g_1,\ldots,g_n,h) \] belongs to \(X_{n+1}\) exactly when \((g_1,\ldots,g_n)\in X_n\) and every \(g_i\) commutes with \(h\). By step 3, the possible first \(n\) coordinates are exactly the elements of \(X_n^h\). Thus the fiber with last coordinate \(h\) has cardinality \(\abs{X_n^h}\), and summing over \(h\) gives the formula.
5. One has\[ \abs{X_{n+1}} = \boxed{\abs G\,\abs{X_n/G}}. \]
Burnside’s lemma applied to the action in step 2 gives \[ \abs{X_n/G} = \frac{1}{\abs G} \sum_{h\in G}\abs{X_n^h}. \] Multiplying by \(\abs G\) and applying step 4 yields the stated identity.
6. Q.E.D.
Step 2 proves the action assertion, and step 5 proves the required counting identity.