Exercise · ZW-BURNSIDE-S-LEMMA-008 · ★★☆ Intermediate
A group of order 3 acts on a set of 7 elements with fixed-point counts 7, 1 and 1. How many orbits are there?
Choose one option.
Show answer
- B. 3
Explanation
The sum is 9 and divided by 3 gives 3 orbits.
Review the concept: Burnside's lemma
Burnside's lemma (also called the Cauchy-Frobenius lemma) states that the number of orbits of a finite group G acting on a set X equals the average number of fixed points of the elements of G.
\[|X/G| = \frac{1}{|G|}\sum_{g\in G}|X^g|\]
More practice problems: burnside's lemma
- ZW-BURNSIDE-S-LEMMA-001 According to Burnside's lemma, what does the expression (1)/(|G|)∑_g|X^g| count?
- ZW-BURNSIDE-S-LEMMA-002 How many elements does the rotation group of the square, C_4, have?
- ZW-BURNSIDE-S-LEMMA-003 Which element of the group always fixes every point of X?
- ZW-BURNSIDE-S-LEMMA-004 If |G| = 6 and the sum of fixed points is 18, how many orbits are there?
- ZW-BURNSIDE-S-LEMMA-005 What alternative name does Burnside's lemma receive?
- ZW-BURNSIDE-S-LEMMA-006 A group of order 4 acts on a set of 10 elements. Its fixed-point counts sum to 4+2+2+0. How many or…
- ZW-BURNSIDE-S-LEMMA-007 How many distinct colorings are there when coloring a square with 3 colors, using only rotations (g…
- ZW-BURNSIDE-S-LEMMA-009 How many distinct necklaces of 4 beads with 2 colors are there, if rotations of C_4 are allowed?