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I'm trying to learn about the Mathieu Groups and I'm reading the following paper by SIMON RUBINSTEIN-SALZEDO.

http://simonrs.com/MathieuGroups.pdf

On the first page it says In order to construct the Steiner system S(5, 6, 12), we will consider the projective line P1(F11) over the field with eleven elements. This consists of the elements 0, 1, 2, . . . , 10, together with an additional element, which we will call ∞. Now consider the set of squares in F11, which are S1 = {0, 1, 3, 4, 5, 9}.

How is this, {0,1,3,4,5,9}, the set of squares in F11? Is there something I'm not understanding? Any help is much appreciated. Thank you. From http://simonrs.com/MathieuGroups.pdf Daniel

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  • $\begingroup$ Identify $F_{11}$ with the integers modulo 11. Square each of them, and see what you get! $\endgroup$ – Gerry Myerson Oct 18 '17 at 2:36
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The set of squares in $F_{11}$ is the set of elements that are some element squared. $\bmod 11$ we have $0^2=0, 1^2=1, 2^2=4, 3^2=9, 4^2=5$ and so on. You can keep going and verify the claim.

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