Problem Set #8 Part B Official Solutions Total Points: 40

Inna Zakharevich

(5) B1. Let H_{g} be the cyclic subgroup of G generated by g ∈ G. Then define the map f : G → {H_{g} : g ∈ G} which maps g to H_{g}. First suppose that for some g ∈ G H_{g} is infinite. Then H_{g} ≅ Z, which has infinitely many subgroups. Thus we see that H has infinitely many subgroups, so G must also have infinitely many subgroups. Next, suppose that G does not have any elements with infinite order; then all H_{g} will be finite. Note that at most |H_{g}| elements map to every H_{g}; thus if there were only finitely many H_{g} G would be finite, a contradiction. So there must be infinitely many H_{g} so G has infinitely many subgroups.

Note that G could have many more subgroups than just these cyclic subgroups.

(5) B2. First note that the number of left cosets of a group H is the same as the number of right cosets. Indeed, we see that if g ∈ aH, so g = ah for some h ∈ H, then g^{−1} = h^{−1}a^{−1} ∈ Ha^{−1}. Thus there is a bijection between the left cosets and the right cosets, so the number of left cosets must be equal to the number of right cosets.

Now suppose that H is a subgroup of G with index 2, and let g ∈ G be such that g ∉ H. Then we know that there are two left cosets: H, gH, and two right cosets, H, Hg^{−1} = Hg since g ∉ H. Since H = H, we see that gH = Hg so H is a normal subgroup.

(15) B3.

(1) (1) Suppose that x, y ∈ Z. We need to show that both xy and x^{−1} are in Z. First note that since xg = gx we know that gx^{−1} = x^{−1}g (by multiplying on the right and left by x^{−1}) so we see that if x commutes with all g then so does x^{−1}. Similarly, if xg = gx and yg = gy we have that gxy = xgy = xyg so xy commutes with g. Thus if x, y ∈ Z then x^{−1}, xy ∈ Z, so Z is a subgroup of G.

(4) (2) Suppose that G/Z is cyclic. That means that there exists a ∈ G such that all cosets are of the form a^{m}Z for some m ∈ Z. Thus every element in G is of the form a^{m}z for some m ∈ Z, z ∈ Z. Fix any x, y ∈ G and write x = a^{m}z_{1} and y = a^{n}z_{2}. Since the z_{i} commute with everything we have

xy = (a^{m}z_{1})(a^{n}z_{2}) = a^{m}a^{n}z_{1}z_{2} = a^{n}a^{m}z_{2}z_{1} = (a^{n}z_{2})(a^{m}z_{1}) = yx

so G is abelian.

(5) (3) Consider a group A, and suppose that #A = p a prime. Then we will show that A is cyclic. Suppose x ∈ A is not the identity. Then the cyclic group generated by x is nontrivial; thus its order must be p and it must be the whole group. Therefore if the order of A is prime then A is cyclic. Now consider G/Z. If [G : H] is prime then G/Z is cyclic; thus if G is nonabelian [G : Z] is not prime and is larger than 1, so [G : Z] ≥ 4.

(5) (4) Fix y ∈ G. Note that C_{y} = {x ∈ G : xy = yx} is a subgroup of G. Thus #C_{y}|#G, so if #C_{y} ≠ #G, #C_{y} ≥ \frac{1}{2}#G. If #C_{y} = #G then y ∈ Z. Thus

#\{(x, y) : xy = yx\} ≤ #Z#G + \frac{1}{2}(#G − #Z)#G = \frac{1}{2}#Z#G + \frac{1}{2}(#G)^{2}.

However, since G is not abelian #Z ≤ \frac{1}{4}#G. So

#\{(x, y) : xy = yx\} ≤ \frac{1}{8}(#G)^{2} + \frac{1}{2}(#G)^{2} = \frac{5}{8}(#G)^{2}.

(10) B4. First note that S_{4} has a subgroup of order 12. First note that S_{4} contains only elements of orders 1, 2, 3, 4; more exactly, it contains 1 element of order 1, 9 of order 2, 8 of order 3 and 6 of order 4. Consider the subset of S_{4} that contains all bijections that fix exactly one element, the identity, and the bijections that switch two disjoint pairs of elements (of which there are 3). This set will contain exactly 12 elements. It also clearly contains the inverses of each of these bijections. Thus to show that it is a subgroup we simply need to show that it contains the compositions of any two of these, which is simple to check (so we will omit it here). We will call this group A_{4}.

Now we will show that A_{4} has no subgroup of order 6. First note that, in a group of order 6 with no element of order 6 (since S_{4} has no element of order 6, A_{4} will not have one either) there will be elements only of orders 1, 2, 3. Note, in addition, that there cannot be more than two elements of order 3. Since elements of order three come in pairs (the element and its inverse) if there are more than two elements of order 3 there will be four such elements. Thus there will only be one element of order 2. Call that element a, and let b be another non-identity element. Since ab ≠ e we know that ab = c where c also has order 3. Also, by process of elimination we see that bc = a. However, then e = a^{2} = abc = c^{2} ≠ e. Contradiction. Thus there cannot be more than two elements of order 3; thus there must be 3 elements of order two. However, since A_{4} contains only three elements of order 2, we see that thus subgroup must contain all three of these elements. The product of any two of these elements will be another element of this form; thus the elements of order 2 (plus the identity) will form a subgroup of order 4 in a group of order 6. Contradiction. Thus A_{4} has no subgroup of order 6.

So now we simply need to show that there will be no other subgroup of order 12. Suppose that we have two subgroups of order 12; call these A_{4} and B_{4}. These will each have one coset. Let X = A_{4} ∩ B_{4}, which will clearly be a subgroup of A_{4} and B_{4}. Let a ∈ A_{4}\X. Then we know that aB_{4} ∪ B_{4} = G. In addition, if for b ∈ B ab ∈ A_{4}, we see that b ∈ A_{4} so b ∈ X. Thus #X = #(A_{4}\X), so #X = 6. Thus A_{4} will have a subgroup of order 6 if there is another group; this is not the case. Thus there is no other subgroup of order 12, and we are done.

(5) B5. First note that if g ∈ aH (for some a ∈ G) then gHg^{−1} = aHa^{−1}. Since g ∈ aH, we know that g = ah for h ∈ H. Thus

gHg^{−1} = (ah)H(ah)^{−1} = ahHh^{−1}a^{−1} = aHa^{−1}.

Let A be the set of all g ∈ G such that if g_{1}, g_{2} ∈ A then g_{1}Hg_{1}^{−1} ≠ g_{2}Hg_{2}^{−1}, so ∪_{g∈G}gHg^{−1} = ∪_{a∈A}aHa^{−1}. We showed above that #A ≤ [G : H]. However, note that each aHa^{−1} will contain e. Thus

#(∪_{a∈A}aHa^{−1}) ≤ 1 + (#A)(#H − 1) ≤ 1 + [G : H](#H − 1) = #G − [G : H] + 1.

This will equal $G if and only if H = G. However, this is not the case, so we know that this will be strictly less than #G, so it cannot be all of G.