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Primer on Group Theory

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Primer on Group Theory Groups eric auld Original: February 5, 2017 Updated: August 15, 2017 Contents What is #Grk;npFpq? I. Group Actions 1.1. Semi-direct products that differ by postcomposition with Aut AutpNq are isomorphic (unfinished). 1.2. Number of subgroups of finite index (unfinished). Maybe has something to do with the notion that G Ñ Sn ö gives isic stabilizers? (if that's true) 1.3. k fixes gH iff k is in gHg´1. kgH “ gH ú g´1kgH “ H g´1kg P H k P gHg´1 1.4. Corollary. The stabilizer of gH in K œ rG : Hs is K X gHg´1. 1.5. Corollary. The fixed cosets of H œ rG : Hs comprise the normalizer of H. The fixed points of H œ rG : Hs are those such that StabH pgHq “ H H X gHg´1 “ H: 1.6. [?, p. 3] If I have a G map X Ñ Y of finite sets, where Y is transitive, then #Y | #X. 1.6.1. Remark. In particular this map must be surjective, because I can get to any y from any other y by multiplication. The action of G œ X is G œ rG : Hs for some finite index subgroup. Look where eH goes...then label Y as G œ rG : Ks where K is the stabilizer of where eH goes. Always Stab'x ¡ Stabx, so we know that K ¡ H. 1 Proof 2: We show that all fibers are the same size. If y1 “ gy, then f ´1rys ÑÐ f ´1ry1s x ÞÑ g ¨ x g´1 ¨ ξ ÞÑ ξ 1.7. [?, p. 23] Normal subgroups play nice with Sylow p subgroups. More explicitly, if P P Sylp G, and N Ÿ G, then P X N P Sylp N and PN{N P SylppG{Nq: In both cases, we show the group is a p group, then show its index isn't divisible by p (so it contains a Sylow p subgroup). Proof of first result: rN : N X P s “ rNP : P s, and the latter isn't divisible by p. Proof of second result: The index thing is easy to see because the index of PN{N in G{N is the index of PN in G, which isn't divisible by p. 1.7.1. Lemma. PN{N is a p group. k Proof 1: Let p P PN{N, then pp “ e, so the order of p divides pk. Proof 2: PN{N – P {N X P . 1.8. [?, p. 3] When N is normal, nppNq | nppGq and nppG{Nq | nppGq. Note that G œ SylppGq and G œ SylppNq both transitive. (For the second, even the action of N on SylppNq is transitive!) Now we have surjective maps of G sets Sylp G Ñ Sylp N Sylp G Ñ SylppG{Nq K ÞÑ K X NK ÞÑ KN{N gKg´1N gKg´1gNg´1 gKNg´1 gKg´1 ÞÑ gpK X Nqg´1 gKg´1 ÞÑ “ “ N N N so we get the divisibility relations on the size of the sets. 1.9. [?, p. 5] A transitive action breaks into equal orbits under a normal subgroup. Given two orbits Nx, Ny, get a set bijection Nx Ñ Ny which is \multiplication by g". nx ÞÑ gnx “ 'pnqgx “ 'pnqy where ' P AutpNq is \conjugation by g". This map is clearly invertible. 2 1.9.1. Example [?, p. 5]. If G ă Sp acts transitively on 1; 2; : : : ; p, then any normal subgroup is either transitive or trivial. ´1 1.10. Stabgx “ g Stabx g . For suppose gx “ y. Then h stabilizes y iff h ¨ pgxq “ gx pg´1hgq ¨ x “ x ´1 So Stabx “ g Staby g. 1.10.1. Example. If G œ X is a finite transitive action, and P is a Sylow p subgroup P P of the stabilizer of x, then NGP acts transitively on X . Let x; y P X . Among those g P sending x y, we look for one in NGP . Since x; y P X , P ă Stabx and P ă Staby.A g sending x y conjugates Stabx to Staby, so it conjugates P ă Stabx to some Sylow p 1 1 P ă Staby. The group P may not be P , but they're conjugate in Staby, so ´1 1 hph “ P h P Staby hP h´1 “ gP g´1 ´1 ´1 ´1 So h g P NGP and h gP g h “ P 1 h´ g : x y 1.11. Corollary. The kernel of a transitive action is the normal core of any stabilizer. For the intersection of all the stabilizers is the kernel, and it's also their normal core, because they're all conjugate. 1.11.1. Example. The kernel of the action G œ rG : Hs is Core H. So if H is normal, H acts trivially on its cosets, and there's a group structure on the cosets. 1.12. Corollary. G œ rG : Hs is equivalent to G œ rG : Ks iff H is conjugate to K. 1.12.1. Lemma. A transitive action G œ X is equivalent to G œ rG : Ks ðñ K is one of the stabilizer subgroups. Therefore, G œ rG : Hs is equivalent to G œ rG : Ks exactly when K is one of the stabilizer subgroups. so if gHg´1 K, then K is the Stab “ g Stab g´1 “ gx x stabilizer of gH. 1.13. Corollary. A subgroup of maximal size is normal. In other words, of index the smallest prime dividing |G|. Let G act on the conjugate subgroups of M by conjugation. The # of conjugate subgroups doesn't exceed the # of cosets, and it divides |G|, so it must be either 1 or the index. If it's the index p, consider what the map G Ñ Sp is. 3 It can only have image size p, because the image size must divide both |Sp| and |G|. Then the kernel (normal core of the stabilizer “ NGH “ H) is the same size as M...X. 1.14. There is a G-map rG : Hs Ñ rG : Ks iff H is conjugate to a subgroup of K. For ´1 look where eH lands, on gK, say. We know H “ StabeH ă StabgK “ gKg . 1.14.1. Remark. Equivalent is: H œ rG : Ks has a fixed point. That is to say, H is a ´1 ´1 subgroup of StabgK for some g. And StabgK “ g StabeK g “ gKg . 1.15. Maps G : H Ñ X pick out H-fixed points of X (unfinished). 1.16. Maps X Ñ rG : Hs where X is transitive pick out H orbits of X with stabilizers inside H (unfinished). See this email https://mail.google.com/mail/u/0/#inbox/ 15dc813436b0a36b 1.17. An action on a finite set X embeds G{ ker into the symmetric group on |X| letters. II. Basic Results 2.1. The centers of GLnpkq and SLnpkq are the scalar matrices. We show that just commuting with the elementary shearing matrices (which are in SLn) forces the matrix to be scalar. a11 a12 1 1 1 1 a11 a12 a a 0 1 0 1 a a ˆ 21 22˙ ˆ ˙ ˆ ˙ ˆ 21 22˙ Add one of column Add one of row 2 1 to column 2 to row 1 a11 a12 ` a11 a11 ` a21 a12 ` a22 a a ` a a a ˆ 21 22 21˙ ˆ 21 22 ˙ This shows that a21 “ 0 and we can do that for any off-diagonal entry. Also, a12 ` a22 “ a12 ` a11 so all diagonal entries are equal. 2.2. [?] The invertible upper triangular matrices are solvable. 2.2.1. When I multiply UT matrices, the diagonal entries just multiply. So the inverse matrix has 1{a, 1{b, etc. on the diagonal. 2.2.2. The commutator is the matrices with ones on the diagonal.. The previous comment implies that a commutator has ones on the diagonal. To show it's the entire commutator, note that pones on diagonalq is a normal subgroup (because diagonal entries multiply), and every right coset has a (unique) representative a diagonal matrix, 1 1 1 a1 a d f a1 1 a d f 1 1 a2 b e a2 1 b e » fi “ » fi » 1 fi a3 c a3 1 c — a ffi — a ffi — 1 ffi — 4ffi — 4ffi — ffi – fl – fl – fl 4 and those representatives form an abelian group pF ˆqn. So the commutator contains these matrices. 2.2.3. When the diagonal is ones, the superdiagonal adds. Because 1 a 1 a1 1 a ` a1 1 b 1 b1 1 b ` b1 » fi » fi “ » fi 1 c 1 c1 1 c ` c1 — 1ffi — 1ffi — 1 ffi — ffi — ffi — ffi – fl – fl – fl plus a bunch of crap on the higher diagonals. 2.2.4. The next commutator is the matrices with zero superdiagonal. The commutator is in there by the previous point. Now we claim every left coset has a unique representative with only a superdiagonal. 1 a d f 1 a 1 d1 f 1 1 b e 1 b 1 e1 “ » 1 cfi » 1 cfi » 1 fi — 1ffi — 1ffi — 1 ffi — ffi — ffi — ffi – fl – fl – fl Since everything is invertible, there's a unique matrix that could be on the right, and it must have zero superdiagonal since superdiagonals add. Therefore every left coset has a unique representative with only a superdiagonal, and the quotient is three copies of the additive group of the field, which is abelian. 2.2.5. When the superdiagonal is zero, the super-superdiagonal adds.
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