Exam #2

The orders of different groups

- cyclic groups (Cn): Cn = n - dihedral groups (Dn): Dn = 2n - Zn = n - For the order of elements, use the formula: Ri ∈ Dn n / gcd (n, i) - Sn = n! - general linear groups (GL2(Zp)): GL2(Zp) = (p2 − 1) · (p2 − p). - alternating group (An): An = n! / 2

What is the order of the permutation (143)(2756)?

3 * 4 = 12

What is the order of the permutation (16345)?

5; count the number of elements in the permutation

How many 2-cycles are there in S6?

6C2 = 15

How do you know if a group is cyclic?

A group G is cyclic if there exists an element "a" in G such that every element of G can be expressed as a power of "a". Mathematically, a group G is cyclic if there exists an element "a" in G such that for every element "g" in G, there exists an integer "n" (positive, negative, or zero) such that: g = a^n In this context, "n" can be any integer, and "a^n" denotes the result of applying the group operation "n" times to the element "a". Every group of prime order is cyclic. e.g. order 11 --> this indicates the group is cyclic.

PART3. For each pair of groups below, either give an isomorphism between them or explain why they are not isomorphic. (c) Z8 and Z2 × Z2 × Z2

Both abelian and order 8. But Z8 is cyclic of order 8 and, in particular, has an element of order 8. But every non-identity element of Z2 × Z2 × Z2 has order 2. Not isomorphic.

PART3. For each pair of groups below, either give an isomorphism between them or explain why they are not isomorphic. (b) D4 and Z4 × Z2

Both have order 8. But still not isomorphic since D4 is non-abelian and Z4 × Z2 is abelian. Non-abelian groups include: dihedral groups, symmetric groups, general linear groups, etc.

How many 5-cycles are there in S7?

For more than 2-cycles, or 3-cycles or more, use the formula for the number of k-cycles = (n choose k) * (k - 1)! 7C5 = 21 * (4)! = 21 * 24 = 504.

PART 1. How many (2, 2, 3)-cycles are there in S7?

General form is (ab)(cd)(efg). Step #1: Start with 7! choices. Step #2: Divide by 2 twice (one for each 2-cycle) because (ab) = (ba) and (cd) = (dc). Divide by 2 again since (ab) and (cd) can be interchanged. Step #3: 3-cycle can be written in three ways. Multiply by 3. 7! / (2^3 * 3) = 210

PART4. What is the order of the permutation (153)(2764) in S7?

Order is the LCM of the orders of the cycles: lcm(3,4) = 12

PART 2. Let H = ⟨R²⟩ as a subgroup of the dihedral group D4 of order 8. That is H = {e, R²} is a subgroup of D4. (b) Show that the quotient D4/H is isomorphic to Z2 × Z2.

Step #1: Determine what D4/H can be represented as. Based on the following cosets, D4/H = {eH, RH, FH, RFH}. Step #2: This suggests an isomorphism. T : D4/H → Z2 × Z2 T(eH) = (0, 0) T(RH) = (1, 0) T(FH) = (0, 1) T(RFH) = (1, 1) A bijection. Step #3: Check for homomorphism. Note a general element of D4/H can be written as RiFjH with i, j ∈ {0, 1}. (refer to guide)

PART 2. Let H = ⟨R²⟩ as a subgroup of the dihedral group D4 of order 8. That is H = {e, R²} is a subgroup of D4. (a) Compute the left cosets of H in D4.

Step #1: H = ⟨R2⟩ contains elements {e, R2} Step #2: Determine the number of left cosets. We know |H| = 2 and D4 = 8. Taking 8/2 = 4, this means there are 4 left cosets and 4 right cosets. Step #3: Identify elements - in this case, e, R, F, and RF. Multiply by elements in H. Step #4: Left cosets of H in D4. Note R is before F. Thus, eH = {e, R2} RH = {R, R3} FH = {F, FR2} = {F, R2F} RFH = {RF, RFR2} = {RF, R3F}

Compute the product (15)(24)(25)(13)(24)(35) and write in standard cycle notation.

Step #1: Start with 1 and go from right to left. 1 --> 3 --> 5 --> 2 --> 4 --> 2 --> 5 --> 1 Step #2: In the following case, 5 --> 2 and 2 --> 5, thus (25). Remaining is 1 --> 3 --> 4 --> 1, thus (134). (134)(25)

PART4. Compute the product (125)(345) in S5.

Step #1: Start with 1 and go from right to left. 1 --> 5 --> 2 --> 1 Step #2: Start with the next number. In this case, 3. 3 --> 5 --> 4 --> 3 Step #3: Write numbers out from right to left in standard cycle notation, taking into account both cycles. (12534)

How many (2,3)-cycles are there in S6?

Step #1: Start with 6! choices. Step #2: Divide by 2 once because (ab) = (ba). Step #3: 3-cycle can be written in three ways. Multiply by 3. 6! / (2 * 3) = 120

PART6. Show that S7 has no subgroup of order 11.

Step #1: Suppose H ⊆ S7 is a subgroup of order 11. NOTE - Since #H is prime (specifically, 11), we know that H is cyclic Step #2: Fix a generator h ∈ H. This must have order 11. The generator has the same order as the order of the group. Step #3: In particular, h ∈ S7 has order 11. But S7 doesn't have any elements of order 11. See below. • e (order 1) • n-cycle (order n), n ≤ 7 • 2, 2-cycles (order 2) • 2, 3-cycles (order 6) • 2, 2, 2-cycles (order 2) • 3, 3-cycles (order 3) • 2, 2, 3-cycles (order 6) • 3, 4-cycles (order 12) • 2, 5-cycles (order 10) 11 does not appear, so no such h exists. Therefore, S7 has no subgroup of order 11.

Compute the product (163)(254) ∘ (17246) and write in standard cycle notation

Step #1: Take the first two groups of parenthesis and write as a permutation. Step #2: Take the last group parenthesis and write as a permutation Step #3: Multiply the two together. Starting from the right permutation, work up to down, and find corresponding letter of the right permutation. Continue to work through. Step #4: Take the product of permutations and then write out in standard cycle notation.

PART5. Consider the function f : R× → R× given by f(x) = x^2 . (c) State the isomorphism coming from f and the first isomorphism theorem.

The domain of f divided by the kernel of f is isomorphic to the image of f. The first isomorphism theorem: R×/ ker f ∼= imf which is R×/{±1} ∼= R+

PART5. Consider the function f : R× → R× given by f(x) = x^2 . (b) Compute the kernel and image of f.

The kernel is the set of all elements which map to the identity, which is 1. ker f = {x ∈ R× ; f(x) = 1} = {x; x^2 = 1} = {±1} ker f = {±1} The image is every value that can apply f(x). Since it is squared, only get positive real values. im f = {f(x); x ∈ R×} = R+ (positive reals) im f = R+

PART3. For each pair of groups below, either give an isomorphism between them or explain why they are not isomorphic. d) Z10 and Z2 × Z5

These are isomorphic. Clearly, Z10 is cyclic. So we show Z2 × Z5 is also cyclic, with generator (1, 1). This is seen to have order 10.

PART3. For each pair of groups below, either give an isomorphism between them or explain why they are not isomorphic. (a) D4 and A4

They have different orders. Thus, not isomorphic since #D4 = 8 and #A4 = 12. *For Dn we know 2n = order of Dn. In this case n = 4, thus (2)(4) = 8. D4 has an order of 8. *For An we know n!/2 = order of An In this case, n! = 4!, thus 24/2 = 12 A4 has an order of 12.

PART5. Consider the function f : R× → R× given by f(x) = x^2 . (a) Show that f is a homomorphism.

fi is homomorphism under multiplication. f(xy) = (xy)^2 = x^2y^2 = f(x)f(y)

What formula should use for 3 or more cycles? What formula should use for 2-cycles?

k-cycles = (n choose k) * (k - 1)! 2-cycles = (n choose k)

How many 3-cycles are there in S6?

k-cycles = (n choose k) * (k - 1)! 6C3 = 20 * (2)! = 20 * 2 = 40.

How many 3-cycles are there in S7?

k-cycles = (n choose k) * (k - 1)! 7C3 = 35 * (2)! = 35 * 2 = 70.