Question

(4) Simplicity II. In this problem, you show that a group G with |G= 30 is not simple. can G have? How many Sylow 3-subgroups? Sylow p-subgroups of G, then either P (a) How many Sylow 5-subgroups (b) Show that, if P and P Pn P(e} (hint: this fact is not true in general, so think again if you didn't use something special to G). (c) If G has the largest possible number of Sylow 5-subgroups, then how many elements of order 5 does G contain? Answer the same (d) Using (4c), show that G is not simple. are or question for p= 3.

Answer 1

A group of prime order is always cyclic.

Solo a. et a be a group of order 30= 2x 3x5 then the order of 2- SSG is a 3-SSG is 3 5-ssh is 5 And The no. of 2-SSG is Hak: + 2k/ 30 - Now I taklis . for f = 0, 1, 2, 7, we have 1+2k/15 .=) no. of 2-SSG are • 1, 3,5,15 lly no. of 3-ssh are 1,10 and no. al .5-5SG are 1,6 (ie. Itsk; 1t58/6) b) cet P, and lq are two sylow p- subgroups of a 2 where olala pipa pa-rope O pis are distinct primes then If Polg = hes then we are done If Popa & he3 al Jaze in a such that aE Popa & ael, and at la a an EP, of an E Pa tn but o(Pi)= prime and 01P2) = prime TT

o So, P, and la are generated by a – Pi = P&. suppose no. of Š SSG are 6 A then no of elements are 4x6 =24 ie. no. of elements of order 5 are 24 Also, we know that a group of even order always . has an element of order a so, atleast one element of order I must O exist so, total elements are 24+1+1=26 But no. of 3-ssa are lilo so, G has atleast two element of order a Also, we know that if a poime divides arder of the group, then a has atleast one element of order So, G has one element Catleast) af order prime, so, G has

and therefore a has unique subgroup of order 3 suppose If a has lo subgroup of order 3 1. then total no. of element of order 3 are ao - no. of sssa isi - a has unique subgroup of order 5 d) by part 'a we see that a has a unique p-sSG I where != 3 or 5. - without loss of genearlity suppose h has unique 5-SSG. say H then o(acha")= o(H) tuto where xhat is onjugate of H but H is unique so, what=H tuta + HOG honce a has a normal subgroup of G. honce G į not simple.