Mathematics 1214: Introduction to Group Theory
Tutorial exercise sheet 7
1. Let n ∈ N, and let α : Z → Z be the mapping α(m) = m + 1 for m ∈ Z.
As usual, we write αk = α
· · ◦ α} if k ∈ N, α0 = ιZ and we also define αk = (α−1 )−k if k ∈ Z
| ◦ ·{z
with k < 0.
n times
(a) Show that αk (m) = m + k for m, k ∈ Z.
(b) Consider the set
G = {αnd : d ∈ Z}.
[In this expression for G, the variable n is fixed, but d is a dummy variable running over Z.
Check that you understand what this means!]
Show that G is a permutation group on Z.
(c) Show that G-orbit equivalence is the same as congruence modulo n.
Solution Let α : Z → Z, n 7→ n + 1. Then α is a bijection. Consider G = {αnk : k ∈ Z}
where α−m = (α−1 )m for m > 0, and α0 = ιZ . It’s not hard to check that G is a subgroup of
(Sym(S), ◦), since it’s clearly non-empty, and αnk ◦ (αnm )−1 = αn(k−m) ∈ G for all k, m ∈ Z.
Let ∼ denote G-orbit equivalence. If a, b ∈ Z then
a ∼ b ⇐⇒ a = β(b) for some β ∈ G
⇐⇒ a = αnk (b) for some k ∈ Z
⇐⇒ a = b + nk for some k ∈ Z
⇐⇒ a − b = nk for some k ∈ Z
⇐⇒ n a − b
⇐⇒ a ≡ b
(mod n).
So ∼ is the same as congruence modulo n.
2. Let a, b, q, r be integers such that a = bq + r. Show that gcd(a, b) = gcd(b, r).
Solution Let d = gcd(a, b). We have d|a and d|b, so d| − bq, so d|a − bq = r. Moreover, if c|b
and c|r then c|bq so c|bq + r = a, so by the definition of d as the gcd of a and b, we have c|d.
In summary:
• d|b and d|r
• if c ∈ Z with c|b and c|r, then c|d.
Hence d = gcd(b, r).
3. It may be shown that if a, b are any integers, then there exist integers s, t such that as + bt =
gcd(a, b).
Use this to prove that if n ∈ N and c, d ∈ Z with gcd(c, n) = 1 then there is a solution x ∈ Z
to the congruence cx ≡ d (mod n).
Solution There exist x, y ∈ Z such that cx + ny = 1. Hence cx = 1 − ny ≡ 1 (mod n).
4. Suppose that n ∈ N and k ∈ Z with n k. Prove that for m ∈ Z we have
n m ⇐⇒ n m + k.
Solution We have n k, so k = nd for some d ∈ Z. Hence
n m ⇐⇒ ∃s ∈ Z : m = ns
⇐⇒ ∃s ∈ Z : m + k = ns + nd = n(s + d)
⇐⇒ ∃t ∈ Z : m + k = nt [reason: we can write t = s + d for ⇒, and s = t − d for ⇐]
⇐⇒ n m + k.
Or, here’s another way, using the fact that ⊕ is a well-defined group operation. Write [s] = [s]n
for s ∈ Z. Then n k, so [k] = [0]. Thus
n m ⇐⇒ [m] = [0] ⇐⇒ [m]⊕[0] = [0] ⇐⇒ [m]⊕[k] = [0] ⇐⇒ [m+k] = [0] ⇐⇒ n m+k.
5. (a) Show that for any real number x and any j ∈ N, we have
xj − 1 = (x − 1)(xj−1 + xj−2 + · · · + x2 + x + 1).
(b) Deduce that if b ∈ Z with b ≥ 2 and j ∈ N, then bj ≡ 1 (mod b − 1).
(c) Show that if n ∈ N0 and a1 , a2 , . . . , an ∈ Z, then
n
n
X
X
j
aj .
aj b ⇐⇒ b − 1 b − 1
j=0
j=0
(d) Is 123456789123456789123456789123456789 divisible by 9?
[Hint: take b = 10 in part (c)].
Solution (a) Multiplying out the brackets on the right-hand side gives:
(x−1)(xj−1 +xj−2 +· · ·+x2 +x+1) = xj +xj−1 +· · ·+x3 +x2 +x−(xj−1 +xj−2 +· · ·+x2 +x+1) = xj −1.
(b) bj − 1 = (b − 1)k where k = bj−1 + · · · + b2 + b + 1, by (a). So b − 1 bj − 1, so bj ≡ 1
(mod b − 1).
P
(c) For each j ∈ {0, 1, 2, . . . , n}, we have b−1 bj −1, so b−1 aj (bj −1), so b−1 nj=0 aj (bj −1).
Hence by Exercise 4,
n
n
n
n
X
X
X
X
j
j
j
aj b −
aj (b − 1) =
aj .
aj b ⇐⇒ b − 1 b − 1
j=0
j=0
j=0
j=0
P35
j
(d) Let N = 123456789123456789123456789123456789. We have N =
j=0 aj · 10 where
a0 = 9, a1 = 8, . . . , a8 = 1, a9 = 9, a10 = 8, . . . , a35 = 1. By (c) with b = 10, we have
P35
P
9 N ⇐⇒ 9 35
j=0 aj . Since
j=0 aj = 4(1 + 2 + · · · + 9) = 4 × 45 = 180 = 9 × 20, we see
P35
aj , so we do have 9 N .
that 9 does divide
j=0
2