Exercise · ZW-ORBIT-STABILIZER-THEOREM-009 · ★★★ Advanced

A group of order \(9\) acts on a set of \(7\) elements. What is the minimum number of fixed points (one-element orbits) there must be?

Choose one option.

Show answer
  • B. \(1\)

Explanation

Orbit sizes divide \(9\) and are at most \(7\): they are \(1\) or \(3\). With \(7=3a+b\) and \(b\) fixed points, \(b=7-3a\equiv1\pmod 3\), so \(b\ge1\) (attained with \(a=2\)).

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Review the concept: Orbit-stabilizer theorem

For a finite group \(G\) acting on \(X\) and an element \(x\in X\), the size of the orbit of \(x\) is the index of its stabilizer in \(G\).

\[|\mathrm{Orb}(x)|=[G:\mathrm{Stab}(x)]=\frac{|G|}{|\mathrm{Stab}(x)|}\]

More practice problems: orbit-stabilizer theorem

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