After Midterm 2

After Midterm 2 - Answer key No.3 to selected homework...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Answer key No.3 to selected homework problems: Math. 110B 8.3 1. Count the number of 4–subsets of the set { 1 , 2 , 3 , 4 , 5 , 6 } : C (6 , 4) = ( 6 4 )= 15. The 4–cycle (1234) generates a cyclic subgroup { I, (1234) , (12)(34) , (4321) } . Another 4–cycle (1324) (made of the same four numbers) generates a different cyclic subgroup { I, (1324) , (13)(24) , (4231) } . Thus one subset of four elements gives rise to at least two cyclic subgroups generated by a 4–cycle. There are 30 of such subgroups. For each four cycle, the product of the 4–cycle and the unique disjoint trasnposition gives an element of order 4. Such element generates again 30 cyclic subgroups; so, there are at least 60 subgroups of order 4. 3. Note | A 4 | = 12. By the third Sylow theorem, the number x of 2–Sylow sub- group is odd and is a factor of 12; so, it has to be either 1 or 3. Since order 4 elements in S 4 are only 4–cycles, which are not in A 4 ; so, A 4 does not have order 4 elements. Thus a Sylow 2–subgroup is made of products of two transpositions: (12) and (34), which is { (12)(34) , (13)(24) , (14)(23) , (1) } . The number y of Sylow 3-subgroup is a factor of 12 and y 1 mod 3. Thus they are either 1 or 4. However one of such ± (123) ² is not normal; so, there are four of them: ± (123) ² , ± (234) ² , ± (134) ² and ± (124) ² . 5. (a) Use the third Sylow theorem (Theorem 8.17). 6. (a) Let G be a group of order 115. 115 = 5 · 23 for primes 5 and 23. The number x of 5–Sylow subgroup is a factor of 115 and x 1 mod 5. The only possibility is x = 1, because factors of 115 are 1 , 5 , 23 and 115 and only 1 satis±es the congruence x 1 mod 5. Thus the Sylow 5–subgroup H is normal (Corollary 8.16). By the same reason, there is only one Sylow 23–subgroup K . Then H = Z 5 and K = Z 23 by Theorem 7.28. Since K H is a subgroup of K and H , it has order which is a common divisor of 5 and 23. Thus K H = { e } .I f kh = k ± h ± ofr k,k ± K and h,h ± H , then K ³ k ± - 1 k = h ± h - 1 H ; so, k ± - 1 k = h ± h - 1 K H = { e } .W e ±nd h ± h - 1 = e h = h ± and similarly, k = k ± . Thus all elements of the form hk are distinct; so, | HK | = | H || K | =5 · 23 = | G | . Thus G = HK . Then by Exercise 25 in Section 8.1, we ±nd that G = H × K = Z 5 × Z 23 . 8. Let G be a ±nite p –group. If a prime q ± ± | G | , then there exists an element of order q by Cauchy’s theorem. Thus q has to be the prime p , because every element of G has p –power order (see the de±nition of p –group in page 255 of the text). Thus p is the only prime factor of | G | ; in other words, | G | = p n for some n> 0. 9. Pick x G . Since G/N is a p –group [ x ] p k =[ e ], where [ x ]= xN = Nx
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This homework help was uploaded on 04/16/2008 for the course MATH 110B taught by Professor Hida during the Winter '08 term at UCLA.

Page1 / 5

After Midterm 2 - Answer key No.3 to selected homework...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online