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College of the Holy Cross, Fall Semester, 2018 MATH 351, Midterm 1 Solutions Thursday, October 4
I. Let G = SL(2, Z), the set of 2 ร 2 integer matrices with determinant 1, which is a group under the operation of matrix multiplication. Let
H =
A =
a b c d
โ G | b โก 0 mod 4
(A) (15) Is H a subgroup of G? Why or why not? Solution: H is a subgroup of G. Hereโs a proof. First, the identity element in G, namely the 2 ร 2 identity matrix:
I =
is in H since the upper right entry is 0 โก 0 mod 4. Next if
A =
a 4 k c d
, and B =
e 4 ` f g
are elements of H, then the product
Aโ^1 B =
d โ 4 k โc a
e 4 ` f g
de โ 4 kf 4 ld โ 4 kg โce + af โ 4 `c + ag
The upper right entry in the product is 4(ld โ kg) and l, d, k, g are all integers, so this element is โก 0 mod 4. Hence Aโ^1 B โ H and H is a subgroup of G. (B) (10) Compute this matrix product, noting that the left and right matrices are inverses of each other and the middle matrix is in H: ( 2 1 3 2
What does the result tell you about the subgroup H? Solution: All the matrices are in G, since they are all integer matrices with determinant equal to 1. The product is ( โ 69 47 โ 116 79
Note that the upper right entry is 47 โก 3 mod 4, not 0 mod 4. Hence this product is not an element of H. It follows that H is not a normal subgroup of G.
II. (A) (15) Let G = ใaใ be a cyclic group. Prove that every subgroup of G is also cyclic. Solution: Let H be any subgroup of G. If H = {e = a^0 } then H = ใeใ, so H is cyclic. Otherwise, let ak^ โ H, where k is the smallest strictly positive power of a for which this holds. Since H is a subgroup, this means that every power of ak^ is also in H, and hence ใakใ โ H. Now to show the other containment, every other element of H is in G, so it has the form an^ for some n. Use division in Z to divide k into n. This means that we can write n = qk + r, where 0 โค r < n. Since an^ โ H and ak^ โ H, ar^ = an^ ยท (ak)โq^ โ H. But by the choice of k, this implies r = 0 and ar^ = e. Hence an^ = (ak)q^ โ ใakใ. Hence since an^ was an arbitrary element of H, we also have H โ ใakใ. Therefore, H = ใakใ is cyclic. This is what we had to show. (B) (5) In part (A), suppose that a has order 120. List all the integers that are orders of elements of G. Solution: The orders are the divisors of 120:
1 , 2 , 3 , 4 , 5 , 6 , 8 , 10 , 12 , 15 , 20 , 24 , 30 , 40 , 60 , 120.
(C) (10) Still assuming a has order 120, how many elements of G have order 24? What are they? Solution: There are ฯ(24) of them. Since 24 = 2^3 ยท3, this number is ฯ(24) = 2^2 ยท(2โ
- ยท (3 โ 1) = 8. To find them, recall that we want the ai^ for which (^) gcd(120^120 ,i) = 24, so gcd(120, i) = 5. The i that work are i = 5, 25 , 35 , 55 , 65 , 85 , 95 , 115.
III. (A) (10) Let H be a subgroup of a group G. Show that aH = bH if and only if aโ^1 b โ H. Solution: โ: Suppose that aH = bH. Then for every h โ H there exists hโฒ^ โ H such that ah = bhโฒ. Multiplying by aโ^1 on the left and hโฒโ^1 on the right we get aโ^1 b = h(hโฒ)โ^1. Since H is a subgroup and h, hโฒ^ โ H, h(hโฒ)โ^1 โ H. Hence aโ^1 b โ H. โ: Suppose that aโ^1 b โ H. Then for all h โ H, we have aโ^1 bh = hโฒ^ โ H. Hence bh = ahโฒ. This implies bH โ aH. Since the left cosets of H in G partition G, this implies that bH = aH. (Recall any two left cosets are either identical or disjoint.) (B) (15) Let G = S 3 and let H =
Find all of the left and right cosets of H in G. Solution: The generator for H is an element of order 2 in S 3. Hence
H =
