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I don't quite understand why Burnside's lemma $$ |X/G|=\frac1{|G|}\sum_{g\in G} |X_g| $$ should be called a "lemma". By "lemma", we should mean there is something coming after it, presumably a theorem. However, I could not find a theorem which requires Burnside as a lemma. In every book I read, the author jumps into calculations using Burnside rather than further theorems.

Question: What are some important consequences of Burnside Lemma, and why is it called a "lemma"?

Ma Joad
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3 Answers3

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Suppose G acts on H where $|G|,|H|<\infty$. For each $g\in G$, consider the permutation character associated with the action. That is, $\chi(g)=|\{\alpha\in H:\alpha\cdot g=\alpha\}|$.
Then, $\sum_{g\in G}\chi(g)=\sum_{\alpha\in H}|G_{\alpha}|=n|G|$, where n is the number of orbits of G on H (the result of Burside's Lemma).

Also, we have the Fundamental Counting Principle: if $O_{\alpha}$ is the orbit of $\alpha$, then $|O_{\alpha}|=|G:G_{\alpha}|$, and since G is finite, $|O_{\alpha}|$ divides $|G|$. Then, as Chris stated, from here we get the class equation.

The number of orbits pops up all over Group Theory, so don't think of Burside's Lemma as a lemma that should have a theorem immediately following, but rather a "tool" that we can refer to very frequently for many theorems.

User7238
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One consequence is for the necklace problem, see this post:

Necklace problem with Burnside's lemma

Math101
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  • I'm confused. The answer to the necklace problem seems to be $\frac{6!}{2!2!2!}$ from a basic calculation. Is the Burnside lemma used to turn an easy calculation into a hard calculation? – Michael Dec 23 '19 at 15:57
  • If you want to answer necklace problems for a generic number of beads (not just a few), you will need Burnside lemma. See for example Proposition 4.1 in this paper: http://dauns01.math.tulane.edu/~vhm/papers_html/par.pdf – Math101 Dec 23 '19 at 18:10
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An closely related fact, by looking at the class equation, is that a nontrivial $p$-group has nontrivial center.