Solutions for Assignment 2
#2 To show that SLn (R) is a subgroup, we need to prove that it is closed under matrix multiplication and contains inverses. If A, B ∈ SLn (R), then det(AB) = det(A) det(B) = 1,
so AB ∈ SLn (R). Likewise, if det(A) = 1, then det(A−1 ) = 1, so SLn (R) is closed
under taking inverses.
Suppose that M ∈ GLn (R). Let
C = SLn (R)M = {AM | A ∈ SLn (R)}
be a right coset of M . Any N ∈ C can be written as a product AM where det(A) = 1,
so det(N ) = det(A) det(M ) = det(M ). Therefore,
C ⊂ {N ∈ GLn (R) | det(N ) = det(M )}
On the other hand, if N ∈ GLn (R) and det(N ) = det(M ), then
det(N M −1 ) = det(N ) det(M )−1 = 1,
so N M −1 ∈ SLn (R) and N M −1 · M = N ∈ C. So
C = {N ∈ GLn (R) | det(N ) = det(M )}.
Thus, there is one coset of SLn (R) for each real number D 6= 0, consisting of all the
matrices with determinant D.
#3 For any g ∈ G, let (Hg)−1 = {x−1 | x ∈ Hg}. We claim that g −1 H = (Hg)−1 .
On one hand, if g −1 h ∈ g −1 H for some h ∈ H, then (g −1 h)−1 = h−1 g ∈ Hg, because
h−1 ∈ H. Therefore, g −1 H ⊂ (Hg)−1 . On the other hand, if (hg)−1 ∈ (Hg)−1 for some
h ∈ H, then (hg)−1 = g −1 h−1 ∈ g −1 H, so (Hg)−1 ⊂ g −1 H, and g −1 H = (Hg)−1 as
desired.
Therefore, g1−1 H = (Hg1 )−1 = (Hg2 )−1 = g2−1 H as desired.
#4
1. The order of any element divides the order of the group, so the order of the group
is a multiple of 3 and a multiple of 7. The least common multiple of 3 and 7 is
21, so the group must have order at least 21.
2. The least common multiple of 6 and 9 is 18, so o(G) ≥ 18. If G = Z18 , a = 3 and
b = 2, then o(a) = 6 and o(b) = 9.
#5 Suppose that n is a multiple of o(a), so that n = ko(a). Then an = ako(a) = (ao(a) )k =
ek = e.
On the other hand, suppose that an = e. Let q, r ∈ Z be such that n = qo(a) + r and
0 ≤ r < o(a). Then
an = aqo(a) ar = ar = e.
Since r < o(a), we know r ≤ 0, so r = 0 and n = qo(a).
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p. 46, #1 Suppose that H and K are subgroups of G. Then H ∩K is nonempty (it contains
e). If a, b ∈ H ∩ K, then ab ∈ H and ab ∈ K (because H and K are closed under
multiplication) so ab ∈ H ∩ K. Likewise, a−1 ∈ H and a−1 ∈ K because H and K are
closed under multiplication, so a−1 ∈ H ∩ K. Therefore H ∩ K is a subgroup.
p. 46, #3 Suppose that a ∈ G is not the identity element. Then hai is a subgroup of G
that is not the identity subgroup. By the assumption, G = hai is a cyclic group and
o(G) = o(a).
Suppose that o(G) = ∞. Then G = {ai | i ∈ Z}, but ha2 i = {a2i | i ∈ Z} is a
nontrivial subgroup of G. This is a contradiction, so o(G) must be finite.
Suppose that o(G) is not prime, and suppose that o(G) = pq, where 1 < p ≤ q < o(G).
Since p < o(G), we have ap 6= e, so hap i = G. But
(ap )q = apq = ao(G) = e,
so o(hap i) ≤ q. This is a contradiction, so o(G) must be finite and prime.
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