This text is intended for a one or two-semester undergraduate course in abstract algebra. Traditionally, these courses have covered the theoretical aspects of groups, rings, and fields. However, with the development of computing in the last several decades, applications that involve abstract algebra and discrete mathematics have become increasingly important, and many sciences, engineering, and computer science students are now electing to minor in mathematics.
PAGES — The content material of the book explains the simple concept of the real numbers of starting. The series and series are elaborated in info and also the diverse techniques and formulas for checking their convergence are mentioned. Those questions are carefully selected in order that the students can practice mathematical knowledge in solving the questions. Though theory still occupies a central role in the subject of abstract algebra and no student should go through such a course without a good notion of what a proof is, the importance of applications such as coding theory and cryptography has grown significantly.
Until recently most abstract algebra texts included few if any applications. However, one of the major problems in teaching an abstract algebra course is that for many students it is their first encounter with an environment that requires them to do rigorous proofs.
Such students often find it hard to see the use of learning to prove theorems and propositions; applied examples help the instructor provide motivation. COM does no longer owns this book neither created nor scanned.
We simply offer the hyperlink already to be had on the internet. If any manner it violates the law or has any troubles then kindly mail us or Contact Us for this hyperlink removal. Thank you.
Download Your Book. Linear Algebra By Samvedna Publication. Your email address will not be published. Prove that G contains an element of order n. Thus, by Question 2. Hence, HN has exactly km elements. Prove that G has a subgroup of order Suppose that N is a normal subgroup of G of order pn.
Prove that G has at least two subgroups of order n. Prove that if H is normal in G, then D is normal in G. Hence, D is normal in G.
Solution: We only need to prove the converse. Thus, H is normal in G. Prove that the intersection of all subgroups of G of order n is a normal subgroup of G. Prove that HK is an Abelian normal subgroup of G. Hence, HK is Abelian. Now, HK is normal by Question 2. Let D be a subgroup of G of order n. Thus, Ord D divides n.
Let K be a subgroup of H. Thus, D is normal in G by Question 2. Solution: We just use a similar argument as in the previous Question. Prove that H is isomorphic to Z. Thus, Ord g divides m. Solution: Since G has no elements of order 2 and 2 is prime, we conclude that 2 does not divide n by Theorem 1. Hence, n is an odd number.
Solution: No. Then by Theorem 1. Then H is cyclic. Hence, H is never cyclic. Prove that H and K are finite groups. Solution: Since G is cyclic, we have H and K are cyclic.
We may assume that H is infinite. By Question 2. Solution: Since Zn is cyclic, by Theorem 1. Since Z24 is cyclic, by Theorem 1.
Solution: NO. Then by Question 2. A Contradiction. Hence, there is no group homomorphism from Z28 onto Z6. Find all elements of Z20 that map to 2, i. Since Z28 is cyclic, Z28 has a unique subgroup of order 7 by Theorem 1. Hence, 5 does not divide Now, by Theorem 1. Hence, there are exactly 10 group homomorphisms from Z30 to Z Prove that G has normal subgroups of index 3 and 5. Solution: Since Z15 is cyclic and both numbers 3, 5 divide 15, Z15 has a subgroup, say, H, of order 3 and it has a subgroup, say, K, of order 5.
Then H is a normal subgroup of Z4 of index 4 and N is a normal subgroup of Z8 of index 4. Prove that G has a normal subgroup of order Solution: Let H be a normal subgroup of Z8 of order 4 and let N be a normal subgroup of Z6 of order 3. Then prove that U 2pn is a cyclic group. Solution: By Theorem 1. Since n U 2 and U p are cyclic groups by Theorem 1. Prove that U pn q m is never a cyclic group.
Prove that up to iso- morphism there are finitely many groups of order n. Solution : Let G be a group of order n. Hence, number of groups of order n up to isomorphism equal number of all subgroups of Sn of order n. Since Sn is a finite group, Sn has finitely many subgroups of order n.
Prove that there is no group homomorphism from Q under addition onto H. Solution: This result is now clear by the previous Question. Let n be a positive integer. Hence, by Question 2. Since S4 is a non- Abelian group, G is non-Abelian.
Solution : Let D be a subgroup of H. The remaining part is now clear. Solution : We know that C H is a subgroup of G. Hence, C H is normal in N H. Hence, H is a normal Sylow p-subgroup of N H. Hence, G is Abelian by Question 2. Prove that G has more than one Sylow 2-subgroup or more than one Sylow 3- subgroup.
Thus, H and K are normal subgroups of G by Theorem 1. A contradiction since G is a non-Abelian group by the hypothesis. Let n be the number of all Sylow 5-subgroups. Thus, H is the only Sylow 5-subgroup of G. By the previous Question, G has a normal subgroup of order 25, say, H. Hence, HK is a subgroup of G by Question 2. Hence, HK is normal by Question 2. Solution: Let n be the number of all Sylow q-subgroups and let m be the number of all Sylow p-subgroups.
Prove that G has exactly 38 elements of order 3. Since H is the only Sylow subgroup of order 19, we have exactly 18 elements in G of order Hence, there are exactly 38 elements in G of order 3. If H is the only Sylow 7-subgroup of G, then by Theorem 1. Since there are exactly 48 elements in G of order 7 and K is a Sylow 2-subgroup of order 8, we conclude that K is the only Sylow 2-subgroup of G. Thus, K is normal by Theorem 1. Prove that G has an element of order Let n be the number of all subgroups of G of order 3.
Let H be the subgroup of G of order 3. Then H is normal by Theorem 1. Hence, there is an element in G of order Let m be the number of all subgroups of G of order 5. Let K be the subgroup of G of order 5. Badawi that G has a subgroup, say, L, of order Thus, as mentioned earlier in the solution G has an element of order Prove that G has exactly one subgroup of order 3 and exactly one subgroup of order 5. Let n be the number of all subgroups of G of order 3, and let m be the number of all subgroups of G of order 5.
Since by the previous Question G has an element of order 15, we conclude by Theorem 1. By an argument similar to the one just given, we will reach to a contradiction. Prove that G has a normal subgroup of order 3 and a normal subgroup of order 5. Prove that G has a normal subgroup of order 6 and a normal subgroup of order 10 and a normal subgroup of order Solution: Let H be a normal subgroup of G of order 2. Prove that G has a subgroup of order 20 and a subgroup of order Solution: By the previous Question G has a normal subgroup of order 10, say, H.
Also, by the previous Question G has a normal subgroup of order 6, say, K. Prove that G has a cyclic subgroup of order 30, that is, show that G has an element of order Solution: Let K be a normal subgroup of G of order 2. Thus, G has a cyclic subgroup of order 30, namely km. Let H be a subgroup of G of order Now, since G has subgroups of order 3 and 5 and 23, G has an element a of order 3 and an element of b of order 5 and an element c of order Since a, b, c commute with each other , by Question 2.
Prove that HK is never a subgroup of G. Prove that H is the only subgroup of G of order p and hence it is normal in G. Solution: Suppose that there is another subgroup, say, K, of G of order p. Thus, H is the only subgroup of order p of G. Thus, H is a Sylow p-subgroup of G. By hypothesis, let K be a normal subgroup of G of order 2.
Show that A Sylow-p-subgroup of G is unique. Solution: Let H be a Sylow-p-subgroup of G. Since G is Abelian, we conclude that H is normal.
Hence H is the only Sylow-p-subgroup of G by Theorem 1. Show that G is Abelian. Solution : Let np be the number of Sylow-p-subgroups and nq be the number of Sylow-q-subgroups. Let H be a Sylow-p-subgroup and K be a Sylow-q-subgroup. Then H and K are both normal in G by Theorem 1. Since q is prime, K is cyclic and hence Abelian. Solution : Let G be a group of order Let n5 be the number of Sylowsubgroups of G. Again by Theorem 1. Let n11 be the number of Sylowsubgroups. Let H be a Sylowsubgroup of G.
Then by Theorems 1. Hence N H is cyclic by Question 2. Thus G has an element of order Let n7 be the number of Sylowsubgroups of G. Thus by Theorems 1. Thus N H is cyclic and hence Abelian by Question 2. Now let K be a subgroup of N H of order 5. Solution : Suppose that A5 has a subgroup H of order 30 20 or Since A5 is non-Abelian simple group see Theorem?? Solution : Let G be a simple group of order Let n5 be the number of Sylowsubgroups, n3 be the number of Sylow- 3-subgroups.
Let H be a Sylowsubgroup. Badawi G has a subgroup of order Now by Theorem 1. Let K be a Sylowsubgroup. Then again by Theorems 1. Thus G has a subgroup of order 6. Let n2 be the number of Sylowsubgroups of G. Thus G is isomorphic to a subgroup of Am by Theorem 1. Thus H contains an element of order 6. Since D is exactly half of H by Question 2.
Thus 15 divides Ord D. Solution : Clearly A5 is a subgroup of S5 of order Let H be a subgroup of S5 of order Hence D is a proper subgroup of A5. Show that G is not simple. Thus assume that G is not Abelian. Then By Theorem 1. Thus Z G is normal in G. Hence G is simple.
Let np be the number of Sylow- p-subgroups of G, nq be the number of Sylow-q-subgroups of G, and nr be the number of Sylow-r-subgroups of G. Since G is simple, by Theorem 1.
Thus G is not simple. Let np be the number of Sylow- p-subgroups of G, nq be the number of Sylow-q-subgroups of G. Hence by Theorem 1. Hence G is not simple. In the first case, since Z45 is cyclic and 15 divides 45, we conclude that G contains an element of order Then G has exactly subgroup of order p by Theorem 1.
Conversely, suppose that G has exactly one subgroup of order p. Then G must be isomorphic to Zpn by Theorem 1.
For G to have exactly one subgroup of order 3, G must have a cyclic a subgroup of order 27 see Question 2. Then clearly that G1 and G2 are nonisomorphic. The subgroup of G1 generated by 0, 9 is cyclic of order 3, and the subgroup of G2 generated by 0, 0, 9 is also cyclic of order 3. Classify G up to isomorphism. Since G has exactly 3 elements of order 2. G can not have a cyclic subgroup of order 8. Thus by Theorem 1. In the first case, G1 has the following elements of order 2, namely : 1, 2, 0 , 1, 0, 0 , 0, 2, 0.
In the second case G2 has the following elements of order 2, namely : 1, 1, 0 , 1, 0, 0 , 0, 1, 0. Show that G has an element of order Also, since 5 divides Ord G , G has an element, say b, of order 5 again by Theorem 1.
Then G is a group of order 8 and hence Ord G is divisible by 4, but each nonidentity element of G is of order 2. Since U 20 is not cyclic, by Theorem 1. But U 4 is isomorphic to Z2 by Theorem 1. How many elements of order 20 does U have?
By Theorems 1. We need to show that m divides n. Show that G has an odd number of elements of order 2. Thus suppose that G is not cyclic. Show that G is cyclic. Thus H has two distinct subgroups of order pi , and thus G has two distinct subgroups of order pi , a contradiction. Hence G is cyclic. What can you say about n. Hence by Question 2. Hence replace p in Question 2. Suppose that G has a normal subgroup H of order 5.
Since G has an odd order, n must be an odd number. Then G has a finite order, say n. Thus G is an infinite group. Thus Ord y divides m, and hence y is of finite order. It is easy to see that Hn is a subgroup of G of order n. Suppose that D is a subgroup of G of order n. Solution : No. Now D has exactly 3 elements of order 3, namely: 2, 2 , 2, 0 , 0, 2.
Show that H is a subgroup of G. Thus H is a subgroup of G by Theorem 1. Solution: First observe that [yx] is the inverse of [xy].
Since y commutes with [xy], we conclude that y commutes with [yx] by Question 2. We prove the claim by induction. Solution: Once again, observe that [yx] is the inverse of [xy]. Since x and y commute with [xy], we conclude that x and y commute with [yx] by Question 2.
Badawi 1. Let L be a subgroup of Z G of order p. Solution 1. Now suppose that G is non-Abelian. Thus H is a normal subgroup of G.
Thus H is cyclic and hence G contains an element of order p2. Solution : Suppose that G is a non-cyclic Abelian. Thus assume that G is non-Abelian. We prove it by induction on n. Now suppose that F is cyclic.
Then G is Abelian by Question 2. Hence F is not cyclic. Then M is a normal subgroup of F. Thus, N is Abelian by Question 2.
If G contains exactly one subgroup of order p, then show that G is cyclic. Thus G contains at least two distinct subgroups of order p, a contradiction. Thus G must be cyclic. If every subgroup of G is normal in G, then show that G is Abelian. Hence G contains two distinct normal subgroups, say H and K, each is of order p. Thus G is Abelian. Prove that H is normal in G. Solution : By Theorem 1. Hence neither p2 divides d nor q divides d, a contradiction. Thus H is normal in G by Question 2.
Solution : First observe that L is a subgroup of K. Also if G is Abelian, then there is nothing to prove. Now let H be a subgroup of G. So we prove the claim for all proper subgroups of G that contain Z G.
0コメント