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In the accepted answer to How many different pentagons in this grid?, Jaap Scherphuis says:

After reducing for symmetry, only 23 are left, which you can check using Burnside's lemma.

How did they get 23 using Burnside's lemma?

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  • $\begingroup$ I really dont get it $\endgroup$ – Sierra Sorongon Jan 27 at 15:44
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Burnside's lemma states that if $G$ is a group acting on set $X$, then the number of orbits $|X/G|$ (i.e. the number of elements of $X$ considered only up to actions by $G$) is given by $$\frac{1}{|G|}\sum_{g\in\ G}|X^g|,$$ where $X^g$ denotes the stabiliser in $X$ of the element $g$ of the group $G$ (i.e. the set of all elements of $X$ fixed by the action of $g$).

For this specific case, $X$ is the set of pentagons and $G$ is the group of the square, i.e. the dihedral group $D_8$ consisting of rotations and reflections. In order to show that the number of pentagons up to symmetry is 23, we need to consider each element of $G$ in turn and find the size of its stabiliser, i.e. the number of pentagons fixed by this rotation or reflection. We sum up all these numbers and divide by 8, and we should get 23.

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  • $\begingroup$ @SierraSorongon: I just answered a different problem involving Burnside's Lemma on the maths stack exchange here. $\endgroup$ – Jaap Scherphuis Aug 23 '17 at 12:27
  • $\begingroup$ Why is it that puzzling.stackexchange.com/questions/10588/… his answer is 26. $\endgroup$ – Sierra Sorongon Aug 23 '17 at 13:00
  • $\begingroup$ @SierraSorongon Huh, good question. I don't fully understand the notation in that answer. Maybe a few of them were counted twice by mistake? $\endgroup$ – Rand al'Thor Aug 23 '17 at 13:06

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