Math 402 Assignment 2. Due Wednesday, October 10, 2012. Chapter 2. Section 4. 1(4.4) Describe All the Groups That Have No Proper

Math 402 Assignment 2. Due Wednesday, October 10, 2012.

Chapter 2. Section 4. 1(4.4) Describe all the groups that have no proper subgroups.

Solution. Let G be a group that contains some element a besides its identity (base) element e. If G has no proper subgroups, then the (cyclic) subgroup generated by a must be the whole group G. That group cannot be inﬁnite because if it were, then a2 would be a proper subgroup. The order n of the ﬁnite group G must be prime: if n were not prime, take a divisor m of n, 1

2(4.5) Prove that every subgroup of a cyclic.

Solution. The inﬁnite cyclic groups can all be realized as Z, +, the group of integers combined by addition. We have seen that the subgroups of Z, + are Zn = {tn|t ∈ Z}, which is the inﬁnite cyclic group generated by n.

Let G be a cyclic group of ﬁnite order n generated by an element a: G = {am|0 ≤ m < n}, where a0 is the base element e. If a subgroup H contains an element am = e, then among the elements of H, consider the element at with the smallest positive exponent t. Then H is the cyclic subgroup of G generated by at. In fact, if H had an element as that was not a power of at, then dividing s by t, we get s = qt + r, with nonzero remainder 0

We can also give the argument for an inﬁnite cyclic group {..., a−n, ..., a−1, e, a, a2, ..., an, ...} in the same spirit. If H is a subgroup containing a power an = e, consider the element at with the smallest positive exponent t. Then you show that at generates H just as you showed it for ﬁnite groups.

3(4.6) (a). Let G be a cyclic group of order 6. How many of its elements generate G? Answer the same question for groups of order 5 and 8.

1 Solution. Present the 6-element cyclic group as {e, a, a2, a3, a4, a5} in terms of a generator a. The only other generator is a5 = a−1.

Present the 5-element cyclic group as {e, a, a2, a3, a4} in terms of a generator a. The other generators are a2, a3, and a4 = a−1, each element of the group is a generator besides e.

Present the 8-element cyclic group as {e, a, a2, ..., a7} in terms of a generator a. The other generators are a3, a5 =(a3)−1, a7 = a−1.

(b). Describe the number of elements that generate a cyclic group of arbitrary order n.

Solution. Let a generate the cyclic group of order n. Then a is an element of order n. The other generators for the group are {am|0 1, then am does not generate the group, i.e., am has order less than n.

Proof of (a). We begin with the general property: if ◦(a)= n, then at = 1 if and only if n|t. Then, consider a power (am)s of am that equals e. By the general property, since ams = 1, n|(ms). Since (n, m) = 1, n|(ms) implies that n|s. Thus, we see that (am)s = e if and only if n|s, and so, am has order n.

Proof of (b). Since m n n m m (a ) (n,m) =(a ) (n,m) = e (n,m) = e, ◦(am) < n.

4(4.7) Let x and y be elements of a group G. If x, y, and xy each has order 2, show that H = {e, x, y, xy} is a subgroup of G.

Solution. H has the inverse of each of its elements since an element of order 2 is its own inverse. Next we need to show that following one element of H by another element of H always produces an element of H. Following y by x gives xy. Since xy has order 2, xyxy = e; following that equation by x gives the equation yxy = x since x2 = e; following the equation yxy = x by y gives the equation xy = yx since y2 = e. Thus, yx = xy. Thus, we see that which element of H preceeds and which follows is immaterial, i.e., if a, b ∈ H, then ba = ab.

2 Using that rule, we see that y(xy) = y(yx) = y2x = x, and (xy)x = (yx)x = yx2 = y. Furthermore, x(xy)= x2y = y, and (xy)y = xy2 = x. Thus, following one element of H by another element of H produces an element of H. Thus, we have shown that H is a subgroup.

5(4.9) How many elements of order 2 does the symmetric group S4 contain?

Solution. We list them: (12), (13), (14), (23), (24), (34), (12)(34), (13)(24), (14)(23). Thus, there are 9 elements of order 2.

6. Show that if a is an element of a group G and if T is a subset of G, then aT = {at|t ∈ T } has the same number of elements as T .

Solution. Deﬁne a mapping φ : T → aT by φ(t) = at for t ∈ T , and deﬁne a mapping τ : aT → T by τ(at) = a−1(at). Then τ ◦ φ = I since τ ◦ φ(t) = τ(at) = a−1(at) = et = t, where e is the identity element of G. In the same way, φ◦τ = I since φ◦τ(at)= φ(a−1(at)) = φ(et)= φ(t)= at. Since φ and τ are inverse mappings, φ : T → aT is a bijection.

Section 5. Problems on homomorphisms and permutations.

7. In the symmetry group of the square, take the closed system (subgroup) consisting of the identity element and the rotation through the angle π, and take the closed system (subgroup) consisting of the identity and the reflection in the line at angle π/4. Show that, when we follow elements in the first system with the elements in the second system, we get a closed system of four elements consisting of the identity, the rotation through angle π, and the reflections in the lines at angles π/4 and 3π/4.

Solution. This problem is a concrete version of problem 4 above. Numbering the vertices 1,2,3,4 as usual, rotation through angle π moves the vertices according to the permutation x = (13)(24), and reﬂection in the line at angle π/4 gives the permutation y = (12)(34). Then xy = (13)(24)(12)(34) = (14)(23).

Let H ⊂ S4 be {e, x, y, xy} = {I, (13)(24), (12)(34), (14)(23)}. By problem 4, H is a subgroup of S4.

8. Find three different subgroups of the symmetry group of the square with 4 elements each, and find five different subgroups with 2 elements each.

Solution. The subgroups of order 4 are {I, (1234), (1234)2, (1234)3}, {I, (13)(24), (12)(34), (14)(23)} from problem 7, and {I, (13), (24), (13)(24)}, which is a subgroup by problem 4.

3 Five subgroups of order 2: {I, (13)(24)}, {I, (12)(34)}, {I, (14)(23)}, {I, (13)}, and {I, (24)}.