Mathematics 351 – Abstract Algebra 1 Solutions for Problem Set 9 November 9, 2007

Section 7.5 5. (a) ℤ_{24} has order 24, so by Lagrange’s Theorem, the possible orders of subgroups of

ℤ_{24} are 1, 2, 3, 4, 6, 8, 12, 24.

(b) S_{4} also has order 4! = 24, so the possible orders of the subgroups of S_{4} are the same as in part (a): 1, 2, 3, 4, 6, 8, 12, 24.

(c) Since |D_{4}| = 8 and |ℤ_{10}| = 10, we have |D_{4} × ℤ_{10}| = 80. Hence the possible orders of subgroups of this group are all the divisors of 80: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80.

6. (a) Let G = (ℤ × ℤ, +), and let H = {(pa, b) : a, b ∈ ℤ}. Then H is a subgroup of G. There only finitely many distinct cosets, namely the cosets

(0, 0) + H, (1, 0) + H, …, (p − 1, 0) + H

(see the solution for problem 12 from Section 6.3 from Problem Set 7). Hence [G : H] = p is finite.

(b) Let G be the same group as in part (a), but now let H = {(0, b) : b ∈ ℤ}. The cosets (a, 0) + H are all distinct. This follows since if a ≠ a′, then (a, 0) − (a′, 0) = (a − a′, 0) ∉ H, so (a, 0) + H ≠ (a′, 0) + H. Hence [G : H] is infinite in this example.

7. The smallest possible order for such a group G is the least common multiple of the integers 1, 2, …, 11, 12. Thinking of the prime factorizations, 2, 3, 4 = 2^{2}, 5, 6 = 2 · 3, 7, 8 = 2^{3}, 9 = 3^{2}, 10 = 5 · 2, 11, 12 = 2^{2} · 3, we see that the least common multiple will be

N = 2^{3} · 3^{2} · 5 · 7 · 11 = 27720.

8. The order n must satisfy n < 100 and 25|n, 10|n. This uniquely determines n = 50 = lcm(25, 10).

9. We are given that H and K are subgroups of G with |H| = |K| = p, p prime. Moreover H ≠ K. We claim first that H ∩ K is a subgroup of H and of K. This follows since H ∩ K is not empty (e ∈ H ∩K). Then a, b ∈ H ∩K implies a, b ∈ H and a, b ∈ K. Hence ab^{−1} ∈ H and ab^{−1} ∈ K since H and K are subgroups. Thus, ab^{−1} ∈ H ∩ K, so H ∩ K is a subgroup of G. Now by Lagrange’s theorem, |H ∩ K| divides |H| = p and |K| = p. But p is assumed prime, so |H ∩ K| is either 1 or p. If |H ∩ K| = p, then H ∩ K = H = K. Since we know that H ≠ K, though, this means that |H ∩ K| = 1. Hence H ∩ K = {e}.

10. By the same fact we used in problem 9 above, H ∩ K is also a subgroup of G, hence of H and K as well. Then by Lagrange’s theorem, we know that |H ∩ K| divides |H| and |K|.

11. We will show first that G must be cyclic, and then deduce that the order of G must be prime. Since G has more than one element, let a ∈ G be any element a ≠ e. Then ⟨a⟩ is a subgroup of G with more than one element. But by hypothesis, G has no non-proper subgroups. This implies that ⟨a⟩ = G, so that G is a cyclic group. We claim next that G must be finite. This follows because if G = ⟨a⟩ was an infinite cyclic group, then G would contain proper subgroups like ⟨a^{2}⟩. This means that G is a finite cyclic group, so G ≃ ℤ_{n} for some n > 1 (Theorem 7.18). Now we claim that n must be prime. If not, then n = km for some integers k, m > 1. But then the order of a^{k} is m < n. This implies that |⟨a^{k}⟩| = m. But then 1 < m < n is the order of a subgroup of G, which contradicts the hypothesis that G has no non-proper subgroups.

13. The order is 30/ gcd(4, 30) = 15. This can be seen directly by computing the powers of a^{4} and using a^{30} = e:

e, a^{4}, a^{8}, a^{12}, a^{16}, a^{20}, a^{24}, a^{28}, a^{32} = a^{2}, a^{6}, a^{10}, a^{14}, a^{18}, a^{22}, a^{26}, a^{30} = e.

This shows that the order of a^{4} is 15, so |⟨a^{4}⟩| = 15 as well. Then the index of ⟨a^{4}⟩ in ⟨a⟩ is |⟨a⟩|/|⟨a^{4}⟩| = 30/15 = 2.

19. If |G| = pq with p, q primes, then the possible orders of subgroups of G are the divisors of pq, namely 1, p, q, pq. The proper subgroups of G have order p or order q. Since both of these are primes, all the proper subgroups of G must be cyclic by Theorem 7.28.

24. We will show this by considering two cases. If |G| = 33 and G has an element a of order 33 (that is, if G is cyclic), then G also has elements of order 3, such as b = a^{11}. (Note that

b = a^{11}, b^{2} = (a^{11})^{2} = a^{22}, b^{3} = (a^{11})^{3} = a^{33} = e.)

Now assume that G is not cyclic (there are no elements of order 33 in G). By Lagrange’s theorem, the possible orders of elements are 1, 3, 11. Suppose that there are no elements of order 3 in G. We will show that this leads to a contradiction, as follows. If there are no elements of order 3, then every element other than e has order 11. Since 11 is prime, by problem 9 above, if a, b are elements of order 11, then either ⟨a⟩ = ⟨ b⟩ or else ⟨a⟩ ∩ ⟨b⟩ = {e}. Take any one element of order 11. We have 11 distinct elements of G in ⟨a⟩. Since there are 33 elements in G altogether, then there must be some b ∈ G with b ∉ ⟨a⟩. But as above, this implies ⟨a⟩ ∩ ⟨b⟩ = {e}. This gives 10 additional elements in G in ⟨b⟩, but not in ⟨a⟩. However, |G| = 33, so there are still elements in G not in ⟨a⟩ ∪ ⟨b⟩. Let c be any one such element. Then ⟨c⟩ ∩ ⟨a⟩ = {e} and ⟨c⟩ ∩ ⟨b⟩ = {e}. Thus ⟨c⟩ contains 10 additional elements of G other than e. But we still have additional elements G since |⟨a⟩ ∪ ⟨b⟩ ∪ ⟨c⟩| = 31. However, now we get a contradiction, since if d is one of the elements of G not in the union of these three subgroups, then d must also have order 11. This is impossible since there are not 10 additional elements in G. Hence G must have some elements of order 3.

Section 7.6

5. (a) G is clearly nonempty. Let A = (\begin{matrix}a & b \\ 0 & d \end{matrix}) and B = (\begin{matrix}a′ & b′ \\ 0 & d′ \end{matrix}) be elements in G. Then

AB^{−1} = (\begin{matrix}a & b \\ 0 & d \end{matrix}) (\begin{matrix}1/a′ & −b′/(a′d′) \\ 0 & 1/d′ \end{matrix}) = (\begin{matrix}a/a′ & (a′b − ab′)/(a′d′) \\ 0 & d/d′ \end{matrix}) .

Since ad ≠ 0 and a′d′ ≠ 0, this matrix is also in G since (a/a′)(d/d′) = (ad)/(a′d′) ≠ 0. This shows that G is a subgroup of GL(2, ℝ), hence a group itself.

Next, Consider N. N is closed under products:

(\begin{matrix}1 & b \\ 0 & 1 \end{matrix}) (\begin{matrix}1 & b′ \\ 0 & 1 \end{matrix}) = (\begin{matrix}1 & b + b′ \\ 0 & 1 \end{matrix}) ∈ N,

and closed under inverses:

(\begin{matrix}1 & b \\ 0 & 1 \end{matrix})^{−1} = (\begin{matrix}1 & −b \\ 0 & 1 \end{matrix}) ∈ N.

Hence N is a subgroup of G.

(b) If A = (\begin{matrix}a & b \\ 0 & d \end{matrix}) ∈ G and B = (\begin{matrix}1 & b′ \\ 0 & 1 \end{matrix}) ∈ N, then

A^{−1}BA = (\begin{matrix}1/a & −b/(ad) \\ 0 & 1/d \end{matrix}) ((\begin{matrix}1 & b′ \\ 0 & 1 \end{matrix}) (\begin{matrix}a & b \\ 0 & d \end{matrix}))
= (\begin{matrix}1/a & −b/(ad) \\ 0 & 1/d \end{matrix}) (\begin{matrix}a & b + b′d \\ 0 & d \end{matrix})
= (\begin{matrix}1 & (b′d)/a \\ 0 & 1 \end{matrix}) .

By definition, this is an element of N. By equivalent statement (2) in Theorem 7.34, this implies that N is a normal subgroup of G.

8. The group table for the quaternion group Q from problem 14 in Section 7.1 is:

1 −1 i −i j −j k −k
1  1 −1 i −i j −j k −k
−1 −1 1 −i i −j j −k k
i  i −i −1 1 k −k −j j
−i  −i i 1 −1 −k k j −j
j  j −j −k k −1 1 i −i
−j  −j j k −k 1 −1 −i i
k  k −k j −j −i i −1 1
−k −k k −j j i −i 1 −1

(a) The cyclic subgroups of Q are:

⟨1⟩ = {1}
⟨−1⟩ = {1, −1}
⟨i⟩ = {1, i, −1, −i} = ⟨−i⟩
⟨j⟩ = {1, j, −1, −j} = ⟨−j⟩
⟨i⟩ = {1, k, −1, −k} = ⟨−k⟩

(b) The cyclic subgroups ⟨i⟩, ⟨j⟩, ⟨k⟩ are all normal by a result we proved in class, since they have order 4 = 8/2. The other two cyclic subgroups are normal since 1, −1 commute with all elements in Q, so aH = Ha if H = {1} or H = {1, −1}.

16. First, K ∩ N is a subgroup of G by the argument given in the solution for problem 9 from Section 7.5 above. Now we will show that K ∩ N is a normal subgroup of G by using equivalent statement (2) from Theorem 7.34. Let g ∈ K ∩ N and a ∈ G. Then g ∈ K and g ∈ N. Since K is normal in G, we have a^{−1}ga ∈ K. Similarly, since N is normal in G, a^{−1}ga ∈ N. Therefore a^{−1}ga ∈ K ∩ N. It follows that a^{−1}(K ∩ N)a ⊂ K ∩ N, so K ∩ N is a normal subgroup of G.

17. We will prove that N ∩ K is normal in K using the equivalent statement (2) of Theorem 7.34. So let g ∈ N ∩ K and let a ∈ K. Then a^{−1}ga ∈ K because a, g ∈ K and K is a subgroup. On the other hand, a^{−1}ga ∈ N since g ∈ N, a ∈ K ⊆ G, and N is a normal subgroup of G. Hence a^{−1}ga ∈ N ∩ K whenever g ∈ N ∩ K and a ∈ K. Therefore N ∩ K is a normal subgroup of K.

22. This follows from the fact proved in Linear Algebra that f(AB) = det(AB) = det(A) det(B) = f(A)f(B).

23. Let A ∈ SL(2, ℝ), so det(A) = 1. If B ∈ GL(2, ℝ) is an arbitrary invertible matrix, then by problem 22,

det(B^{−1}AB) = \frac{1}{det(B)} det(A) det(B) = det(A) = 1.

Hence for all B,

B^{−1}SL(2, ℝ)B ⊆ SL(2, ℝ).

Hence SL(2, ℝ) is a normal subgroup of GL(2, ℝ) by Theorem 7.34 (equivalent statement (2)).