Question

Proposition 6.4. There are (up to isomorphism) exactly three di groups of order 12: the dihedral group De, the alternating group A, and a generated by elements a,b such that lal 6, b a', and ba a-b. stinct nonabelian SKETCH OF PROOF. Verify that there is a group T of order 12 as stated (Exercise 5) and that no two of Di,A,T are isomorphic (Exercise 6). If G is a non- G: P) - 4. By Proposition 4.8 there is a homomorphism f G S, whose kernel K is contained in P, whence K - Por (e). If K- (e), fis a monomorphism and G is isomorphic to a subgroup or oraer 12 of S, which must be As by Theorem Otherwise K-P and Pis normal in G. In this case P is the unquesylow 3-subgroup. Hence G contains abelian group of order 12, let P be a Sylow 3-subgroup of G. Then |P 3 and elements of order 3. If c is one of these, then G: Cole)l-1 or 2 since (G: Colc)l is the number of conjugates of c and every con- jugate of e has order 3. Hence Cofe) is a group of order 12 or 6. In either case is de Coe) of order 2 by Cauchy's Theorem. Verify that led 6. there Let a- ed; then (a) is normal in G and |G/lal-2. Hence there is an element bs G such that b İ (a), b a-6, baba is the only possibili possibilities for b, (a). У." aaotP-a. lead to c e, b, (a), and bab-le (a). Since G is nonabelian a y; that is, ba-a-b. There are s contradictions; b -12 and G abelian. Therefore, the only possibilities are a or b 6;b-e, ba-ar'b, whence G Da by Theorem 1.6.13 -6,b-a ba- ab, whence GT by Exe Gii) lal The table below lists f
please look at red line

please explain why P is normal

thanks

Answer 1

f:G--,s, homomorphism ίδα and as k is contawed in P, and P 13 pnme P does not hoe prr uoup Hence, either k= < e > k=P07 P is horma becau sep and kis the reinel of the неосе р is normal aJkep