Download Subgroup - Abstract Algebra - Solved Exam and more Exams Algebra in PDF only on Docsity!
Answers to Final Exam
MA441: Algebraic Structures I
20 December 2003
- Definitions (20 points)
- Given a subgroup H C G, define the quotient group G/H. (Describe the set and the group operation.) The quotient group is the set of left (or right) cosets {aH|a ∈ G} with group operation (aH)(bH) = (ab)H.
- Given a permutation group G < Sn acting on the set { 1 , 2 ,... , n}, define the stabilizer StabG(i). StabG(i) = {φ ∈ G|φ(i) = i}
- Given an element a ∈ G, define the centralizer C(a). C(a) = {g ∈ G|ag = ga}
- Given a ∈ G, define the conjugacy class cl(a). cl(a) = {gag−^1 |g ∈ G}
- Fill in the blanks or answer True/False (five from this list)
- True or False: (1234)(4567) ∈ A 7. True A 4-cycle is odd (a product of three 2-cycles), and two odds make an even.
- True or False: 〈(14)〉 is a normal subgroup of S 4. False For example, (12)(14)(12) = (1)(24), which is not contained in 〈(14)〉.
- True or False: If 7 divides |G|, then G has an element of order 7. True This follows from Cauchy’s Theorem.
- True or False: For every positive integer n, Aut(Zn) ≈ U (n). True This is Theorem 6.5.
- True or False: Let G be a cyclic group of order n. If k|n, then there is an H < G such that H has order k. True This follows from the Fundamental Theorem of Cyclic Groups (Theo- rem 4.3).
Let H be a nonempty finite subset of a group G. Prove that H is a subgroup of G if H is closed under the operation of G. This is Theorem 3.3, the Finite Subgroup Test. One need only prove that any element a ∈ H has an inverse. If a = e, then a is its own inverse. Assume a 6 = e. Consider 〈a〉. Since H is finite, by the Pigeonhole Principle, there are i, j such that ai^ = aj^. We may assume that 0 < i < j. Then aj−i^ = e and a · aj−i−^1 = e. Since a 6 = e, j − i − 1 > 0. Therefore a−^1 = aj−i−^1.
Use Lagrange’s Theorem to prove Fermat’s Little Theorem: for every integer a and every prime p, ap^ ≡ a (mod p). This is Corollary 5 to Lagrange’s Theorem, Theorem 7.1. Apply La- grange’s Theorem to U (p), which has order p−1. Because ap−^1 ≡ 1 (mod p), we multiply both sides by a to get ap^ ≡ a (mod p).
Cosets
- Given a subgroup H < G and any a, b ∈ G, prove that either aH = bH or aH ∩ bH = ∅, i.e., aH and bH are disjoint. This is from the Lemma on page 135 of Chapter 7. Suppose x is in both aH and bH. Then we can write x = ah 1 = bh 2 for some h 1 , h 2 ∈ H. Therefore a = bh 2 h− 1 1 and b = ah 1 h− 2 1. So aH = bh 2 h− 1 1 H ⊆ bH and conversely bH = ah 1 h− 2 1 H ⊆ aH. Therefore aH = bH. Alternatively, one could cite the property that aH = H iff a ∈ H. Then aH = b(h 2 h− 1 1 H) = bH, since h 2 h− 1 1 H = H.
We calculate
28 = 2 · 13 + 2 13 = 6 · 2 + 1 1 = 13 − 6 · 2 2 = 28 − 2 · 13 1 = 13 − 6(28 − 2 · 13) = 13 · 13 − 6 · 28.
Reducing the last equation modulo 28, we see that 13 is its own inverse in U (28).
Prove that if |a| = k, then 〈a〉 = {e, a, a^2 ,... , ak−^1 }. This is Theorem 4.1. Note that 〈a〉 = {an|n ∈ Z}. The key idea here is to apply the Division Algorithm. For any n, we can divide by k and take the remainder to get n = q · k + r, where 0 ≤ r < k. Then an^ = ar. (For n < 0, the q will be negative, and this still works.)
Let G be the group {a + b
2 | a, b ∈ Q} under addition, and let H be the group
H =
{[
a 2 b b a
] ∣∣
∣ a, b^ ∈^ Q
under addition. Show that G and H are isomorphic under addition. This was problem 6.24 in Homework Assignment 9. Let φ be the map such that
φ(a + b
a 2 b b a
The map φ is one-to-one because if φ(a + b
- = φ(c + d
2), then ( a 2 b b a
c 2 d d c
then the corresponding entries are equal, so a = c and b = d. The map φ is onto because for any a, b, the definition of φ above means a + b
2 is a preimage for (^) ( a 2 b b a
under φ. To show the homomorphism property, observe that ( a 2 b b a
c 2 d d c
a + c 2(b + d) b + d a + c
So therefore
φ((a + b
2)) = φ((a + c) + (b + d)
- = φ(a + b
- Suppose that |x| = n. Find a necessary and sufficient condition on r and s such that 〈xr〉 ⊆ 〈xs〉. Justify your answer.
This was problem 4.60 in Homework Assignment 7. The proof we followed in the homework was to apply Theorem 4.2 to get
〈xr〉 = 〈xgcd(r,n)〉, 〈xs〉 = 〈xgcd(s,n)〉,
and then deduce the divisibility condition gcd(s, n) divides gcd(r, n). Alternatively, one could argue that by the Fundamental Theorem of Cyclic Groups, 〈xr〉 ⊆ 〈xs〉 iff |〈xr〉| divides |〈xs〉|. Since 〈xs〉 is a cyclic group, there is exactly one subgroup for each order dividing |〈xs〉|.
- (for 10 points of extra credit) Prove Lagrange’s Theorem. You may cite basic properties of cosets, such as those listed in Gallian’s Lemma in Chapter 7, if you state them accurately. Please refer to the proof on page 137 of Gallian. It suffices to note that G can be partitioned by its cosets, that all cosets have the same size, and that therefore the order of G is an even multiple of the order of a coset.
