21-373: Algebraic Structures Assignment 2

Peter Battaglino 3. February, 2005

Problem 3.26: Suppose n is an even positive integer and H is a subgroup of ℤ_{n}. Prove that either every member of H is even or exactly half of the members of H are even.

\underline{Solution:} Since ℤ_{n} is a cyclic group, we know that all of its subgroups have to be cyclic. Since H is therefore cyclic, we can write H = ⟨a⟩ for some a ∈ ℤ_{n}. Suppose a is even. Then ∑_{i=1}^{k} a must be even for any positive integer k, and all of the elements of ⟨a⟩ = H must be even. Now suppose a is odd. Then the elements of {a, a + a, a + a + a, ⋯, 0} alternate between being even and odd as we traverse them in order. If this set has an even number of elements (which are necessarily distinct by definition of the group generated by a), then we see that half of them are even and half of them are odd, which was to be shown. To show that the order of ⟨a⟩ must be even, given that n is even, we observe that |a|a mod n = 0. Since a is odd by assumption, we conclude that |a| must be even if |a|a is to be divisible by an even number (no odd number is divisible by an even number). Thus we have that the order of the group generated by an odd a must be even, so half the elements are even and half the elements are odd.

Problem 3.52: Let G be a finite group with more than one element. Show that G has an element of prime order.

\underline{Solution:} Let’s pick some element g ∈ G that has order |g| = m. Suppose m is not a prime number, and is not equal to 1. m has the prime decomposition

m = p_{1}^{q_{1}}p_{2}^{q_{2}} ⋯ p_{n}^{q_{n}},  (1)

where each of the q_{i} are positive integers and each of the p_{i} are prime. Since g has order m, we know that

g^{p_{1}^{q_{1}}···p_{n}^{qn}} = 1.  (2)

We can use the properties of exponents to write

(g^{p_{1}^{q_{1}}···p_{n}^{q_{n−1}}})^{p_{n}} = 1,  (3)

which says that the order of the element g′ = g^{p_{1}^{q_{1}}···p_{n}^{q_{n−1}}} divides p_{n}. Since p_{n} is a prime number, we know that either |g′| = 1 or |g′| = p_{n}. The case |g′| = 1 contradicts the assumption that |g| = m, so we take |g′| = p_{n}. Thus we have found an element |g′| ∈ G with prime order, given any element g with non-prime (and non-unity) order.

Battaglino 3 Problem 4.32: Determine the subgroup lattice for ℤ_{12}. \underline{Solution:} We can list the subgroups of ℤ_{12} as

⟨0⟩ = {0}
⟨1⟩ = ℤ_{12}
⟨2⟩ = {0, 2, 4, 6, 8, 10}
⟨3⟩ = {0, 3, 6, 9}
⟨4⟩ = {0, 4, 8}
⟨5⟩ = ℤ_{12}
⟨6⟩ = {0, 6}
⟨7⟩ = ℤ_{12}
⟨8⟩ = {0, 4, 8} = ⟨4⟩
⟨9⟩ = {0, 3, 6, 9} = ⟨3⟩
⟨10⟩ = {0, 2, 4, 6, 8, 10} = ⟨2⟩
⟨11⟩ = ℤ_{12}

The subgroup lattice is illustrated in Fig. 1.

Battaglino 4 Figure 1: The subgroup lattice for ℤ_{12} Problem 4.47: Determine the orders of the elements of D_{33} and how many there are of each.

\underline{Solution:} Let R = R_{2π/33}. The group D_{33} contains 33 reflections and 33 rotations (the identity being included in the rotations). Each of the 33 reflections has order 2. The identity rotation is of order 1. Since the only divisors of 33 are itself, 1, 11, and 3, we know that all the subgroups of the group generated by R must have order of 1, 3, 11, or 33. All the groups with order 11 satisfy 11 = 33/ gcd(33, k), where R^{k} is the generator of the group. The values of k which satisfy this equation are 3, 6, 9, 12, 15, 18, 21, 24, 27, and 30. Thus there are 10 groups with order 11. The groups of order 3 satisfy gcd(33, k) = 11, which has the solutions 11 and 22. The rest of the elements in the subgroup ⟨R⟩ ⩽ D_{33} must all have order 33, so there are 20 elements with order 33.

To summarize, there is one element with order 1, there are 33 elements with order 2, 2 of order 3, 10 of order 11, and 20 of order 33.

Problem “Additional Problem”: Let H, H′ be subgroups of a group G. Prove that H ∪ H′ is a subgroup of G if and only if one of the subgroups H, H′ is a subset of the other.

\underline{Solution:} ⇒: Let a ∈ H, b ∈ H′, and a ∉ H′. Since a, b ∈ H ∪ H′ and H ∪ H′ is a group, we know that ab ∈ H ∪ H′, which implies that ab ∈ H or ab ∈ H′. Suppose ab ∈ H. Then, since a ∈ H and H is a group, we know that a^{−1} ∈ H and thus that a^{−1}ab ∈ H, which means that b ∈ H. Now suppose that ab ∈ H′ instead. Then abb^{−1} ∈ H′, so a ∈ H′, which contradicts the assumption a ∉ H′. Thus we see that if we take one element of H and one element of H′, and restrict one of the elements to lie in one of H, H′ but not the other, then the other element must be a member of the same subgroup as the first, which means that either H ⊂ H′ or H ⊃ H′ since our argument is valid for arbitrary a and b. ⇐: Let H ⊃ H′. Then H ∪ H′ = H, and since H is a group, H ∪ H′ is also a group. Let H ⊂ H′. Then H ∪ H′ = H′, and since H′ is a group, H ∪ H′ is also a group.