Math 4400/6400 – Homework #5 Solutions
MATH 4400 problems.
1. Since 43 is prime, we know that the group U (Z43 ) has a generator. Find a positive
integer a for which [a] is a generator (and prove that a really is a generator!).
Solution. By Fermat’s little theorem, 342 ≡ 1 (mod 43), and so 3 has order dividing
42. We have to rule out that the order of 3 is a proper divisor of 42. Note that the
divisors of 42 are 1, 2, 3, 6, 7, 14, 21, 42.
We compute that modulo 43,
31 ≡ 3,
32 ≡ 9,
34 ≡ 81 ≡ −5,
38 ≡ (−5)2 ≡ 25,
316 ≡ 252 = 625 ≡ −20.
Using this, we can check that modulo 43,
321 ≡ 316 · 34 · 3 ≡ (−20) · (−5) · 3 ≡ 100 · 3 ≡ 14 · 3 = 42 ≡ −1
(mod 43).
Since 321 6≡ 1 (mod 43), the order is not a divisor of 21, which rules out 1, 3, 7, 21 as
possibilities. We also compute that
314 = 38 · 34 · 32 ≡ 25 · (−5) · 9 ≡ 25 · (−45) ≡ 25 · (−2) ≡ −50 ≡ −7
(mod 43).
It follows that the order is not a divisor of 14, which further rules out 2 and 14.
Finally,
36 = 32 · 34 ≡ 9 · (−5) = −45 ≡ −2
(mod 43).
So the order is not a divisor of 6. This further rules out 6.
The only remaining possibility is that 3 has order 42, as desired.
2. Let H be a subgroup of the abelian group G. Recall that this means that H becomes
a group if we ‘import’ the multiplication operation from G. Show that the identity
element of H is the same as the identity element of G.
Proof. Let 1H be the identity of H and let 1G be the identity of G. Since G is a group,
there is a unique z ∈ G with
1H ⊗ z = 1H .
Now z = 1H satisfies this equation, since 1H is an identity element for H. But z = 1G
also satisfies this equation, since 1G is an identity element for all of G. So 1H = 1G .
Alternative proof. Let’s forget about H for a second. Suppose that x ∈ G is an element
which satisfies x ⊗ x = x. Then x−1 ⊗ (x ⊗ x) = x−1 ⊗ x. But the left hand side is
1G ⊗ x = x, and the right hand side is 1G . So x = 1G .
Now we get back to the problem. Observe that 1H satisfies 1H ⊗ 1H = 1H , since
1H is an identity element for H. Thus, the result of the last paragraph implies that
1H = 1G .
3. Let H be a fixed subgroup of G. For each x ∈ G, we define a subset xH ⊂ G by
xH = {x ⊗ h : h ∈ H}.
[This is called the coset of x in G, with respect to H.]
a. Show that if x, y ∈ G, then either xH = yH or xH and yH are disjoint sets.
Proof. We suppose that xH is not disjoint from yH and we show that then xH =
yH. Assume that z ∈ xH ∩ yH. Then z can be written as x ⊗ h1 for some h1 ∈ H
and also as y ⊗ h2 for some h2 ∈ H. So x ⊗ h1 = y ⊗ h2 , which implies that
−1
x = y ⊗ h2 ⊗ h−1
1 . Putting h3 = h2 ⊗ h1 , we can write x = y ⊗ h3 , where h3 ∈ H.
Every element of xH is of the form x ⊗ h. Observe that x ⊗ h = y ⊗ (h3 ⊗ h).
Since h3 ⊗ h ∈ H, we see that x ⊗ h ∈ yH. So xH ⊂ yH.
On the other hand, if x = y ⊗ h3 , then y = x ⊗ h4 , where h4 = h−1
3 ∈ H. Every
element of yH has the shape y ⊗ h. Since y ⊗ h = x ⊗ (h4 ⊗ h) ∈ xH, it follows
that yH ⊂ xH. So yH = xH.
b. If G is finite, show that every coset has the same size as H. Deduce that there
#G
distinct cosets.
are precisely #H
Proof. We exhibit a bijection between H and xH for all x ∈ H. Let φ : H → xH
be the map h 7→ x ⊗ h. φ is a bijection, because it has an inverse map, namely
the map ψ : xH → H given by ψ(g) = x−1 ⊗ g. So all cosets have the same size
as H.
Finally, notice that all elements of G belong to some coset, since x is always an
element of xH. It follows that G is the disjoint union of its distinct cosets. Since
#G
.
each of the cosets has size H, the number of distinct cosets must be #H
c. If G = Z (considered as a group under addition) and H = mZ for a positive
integer m, how many different cosets of H in G are there? What are they?
Proof. The cosets of mZ are 0 + mZ, 1 + mZ, . . . , (m − 1) + mZ, which consist
(respectively) of the numbers congruent to 0, 1, 2, . . . , m − 1 modulo m.
4.
a. Let k and m be integers, with m > 0. Show that the order of [k] in Zm (viewed as
a group under addition) is m/ gcd(k, m). Hint: The case when k = 4 and m = 30
was worked out in class.
n times
z
}|
{
Proof. The order of [k] in Zm is the least integer n > 0 for which [k] + · · · + [k] =
[0], i.e., the least n for which kn ≡ 0 (mod m). Let d = gcd(k, m). Then, from
our big theorem on linear congruences,
kn ≡ 0
(mod m) ⇐⇒
k
n≡0
d
(mod
Moreover,
k
n≡0
d
(mod
2
m
m k
) ⇐⇒
| n.
d
d d
m
).
d
But gcd(m/d, k/d) = 1. So by the fundamental lemma, md | kd n precisely when
m
| n. Finally, we make the trivial observation that the least positive integer n
d
for which md | n is n = m/d itself.
b. Now let G be an arbitrary abelian group. Suppose that x ∈ G is an element of
order m. Show that for each k ∈ Z, the element xk has order m/ gcd(m, k).
Proof. The order of xk is the least integer n for which xkn = 1. Since x has order
m, we know that xkn = 1 if and only if m | kn. So we are seeking the least integer
n > 0 for which m | kn. But this is exactly the problem we solved in part a. –
the least such n is m/ gcd(m, k).
Remark: Part b. is a generalization of a., which corresponds to the case when
G = Zm and x = [1].
5. Let p be an odd prime. In the last HW, you showed that if p ≡ 1 (mod 4), then −1
!)2 ≡ −1 modulo p. In this exercise, we
is a square modulo p, and in fact that ( p−1
2
explore the converse.
Suppose that x2 ≡ −1 (mod p) for an x ∈ Z. Show that [x] is an element of order
4 in U (Zp ), and deduce that p ≡ 1 (mod 4). [Combined with your work on your
previous HW, this shows that among odd primes p, the ones for which −1 is a square
are precisely those p ≡ 1 (mod 4).]
Proof. If x2 ≡ −1 (mod p), then [x]4 = [x4 ] = [(−1)2 ] = [1] in U (Zp ), and so [x] has
order dividing 4. Since [x]2 is [−1] and not [1] in U (Zp ), the order cannot be 1 or 2. So
[x] has order exactly 4. But the order of x divides #U (Zp ) = p − 1. So p ≡ 1 (mod 4),
as desired.
6.
a. Suppose that G is a cyclic group of size n. Show that for every divisor d of n,
there are φ(d) elements of G of order d.
Proof. Fix a generator x of G (equivalently, an x ∈ G of order n). The hint
reduces the problem to counting the number of m ∈ {0, 1, 2, . . . , n − 1} for which
xm has order d. By problem 3b., the order of xm is n/ gcd(m, n), so for xm to
have order d, we need gcd(m, n) = n/d.
In particular, we must have that nd is a divisor of m, so that m = nd k. Since
0 ≤ m < n, we need 0 ≤ k < d. Moreover, for such m,
gcd(m, n) =
n
n
n
⇐⇒ gcd( k, n) =
d
d
d
n
n
⇐⇒ gcd(k, d) = ⇐⇒ gcd(k, d) = 1.
d
d
So we have reduced the problem to counting the number of integers k with 0 ≤
k < d and gcd(k, d) = 1. But this is exactly φ(d).
b. Prove that for every positive integer n, we have
X
φ(d) = n.
d|n
3
Proof. Let G be the group Zn under addition, which is cyclic of order n. Then
every element of G has a unique order, and that order is a divisor of n. For each
divisor d of n, there are precisely φ(d) elementsPof G of order n, by part a. Since
there are n elements total of G, we must have d|n φ(d) = n, as asserted.
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MATH 6400 problems. Do any two of the following four.
G1. If you have done problem 5 above, then you have characterized the primes p for which
−1 shows up in the list of squares modulo p. In this exercise, we will characterize those
primes p for which −3 is a square mod p. We suppose throughout that p > 3 is prime.
a. Show that if p ≡ 1 (mod 3), then −3 belongs to the list of squares modulo p.
Proof. Since U (Zp ) is cyclic, there is an element of order p − 1. As shown in
class, this means that there is an element of every order dividing p − 1, and so
in particular there is an element of order 3. Call this element [a]. Then a3 ≡ 1
(mod p), so that p | a3 − 1 = (a − 1)(a2 + a + 1).
It cannot be that p | a − 1, as otherwise a ≡ 1 (mod p) and then a would have
order 1, not 3. So p | a2 + a + 1. In other words, a2 + a + 1 ≡ 0 (mod p).
Multiplying through by 4, we find that
4a2 +4a+4 ≡ 0
(mod p),
and so (2a+1)2 = (4a2 +4a+4)−3 ≡ −3
(mod p).
So −3 is on the list of squares modulo p.
b. Show that if there is a square root of −3 modulo p, then p ≡ 1 (mod 3). Hint:
Use the square root of −3 to construct an element of U (Zp ) of order 3.
Proof. We reverse engineer the proof of a. Suppose x2 ≡ −3 (mod p), and choose
an integer a so that 2a ≡ x − 1 (mod p). (Since the gcd of 2 and p is 1, this
congruence is solvable!) We claim that a is an element of order 3 modulo p.
First, we check that a2 + a + 1 ≡ 0 (mod p). Since 4 is coprime to p, this is
equivalent to showing that 4(a2 + a + 1) is a multiple of p. We have
4a2 ≡ (x − 1)2 = x2 − 2x + 1 ≡ −2x − 2
(mod p),
while
4a + 4 ≡ 2(x − 1) + 4 ≡ 2x + 2
(mod p).
Hence,
4a2 + 4a + 4 ≡ (2x + 2) − 2x − 2 ≡ 0
(mod p).
Now that we know p | a2 +a+1, it follows that p | (a2 +a+1)(a−1) = a3 −1, so that
a3 ≡ 1 (mod p) and ordp (a) | 3. Thus, ordp (a) = 1 or ordp (a) = 3. If ordp (a) = 1,
then a ≡ 1 (mod p). But then a2 + a + 1 ≡ 3 (mod p). Since p | a2 + a + 1, this
forces p to divide 3, so p = 3, contrary to our opening assumption that p > 3.
So we have constructed an integer a which has order 3 mod p. Since every order
divides p − 1, it must be that p ≡ 1 (mod 3).
G2. According to Fermat’s little theorem, the following statement holds for every prime p:
ap−1 ≡ 1
(mod p) for every a ∈ Z relatively prime to p.
(**)
Show that p = 561 has the property (**), even though 561 = 3 · 11 · 17 is not a prime
number.
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Proof. If a is coprime to 561, then a is coprime to each of 3, 11, and 17. So by Fermat’s
little theorem, a2 ≡ 1 (mod 3), a10 ≡ 1 (mod 11), and a16 ≡ 1 (mod 17). All of 2, 10,
and 16 divide 560. It follows that a560 ≡ 1 modulo each of 3, 11, and 17. In other
words, all of 3, 11, and 17 divide a560 − 1. By unique factorization, it follows that
561 = 3 · 11 · 17 | a560 − 1, i.e., that a560 ≡ 1 (mod 561).
G3. If p is prime, then (p − 1)! ≡ −1 (mod p), as you showed in your last HW. Show that
if n is composite, then (n − 1)! ≡ 0 (mod n) with precisely one exception, and find
that exception.
Proof. The only exception is n = 4, where (4 − 1)! = 6 ≡ 2 (mod 4).
We have to show that if n is composite and n 6= 4, then n | (n − 1)!. Let a be the
smallest divisor of n that is larger than 1. It is easy to see that the minimality of a
forces a to be prime. Let b = n/a. If a 6= b, then both a and b appear in the list
1, 2, 3, . . . , n − 1, and so n = ab | 1 · 2 · · · · · (n − 1) = (n − 1)!, as desired.
So the only possible exceptions correspond to the case when b = a. In that case, n = a2
and (as already noted) a is prime. If a > 2, then the list 1, 2, 3, . . . , n − 1 = a2 − 1
contains both a and 2a. Thus,
n = a2 | a · (2a) | 1 · 2 · · · · · (n − 1) = (n − 1)!.
So the only remaining possible exception corresponds to when a = 2, when n = a2 = 4.
We have already noted this exception above.
G4. Let p be an odd prime, and suppose that k is an integer with 1 ≤ k < p − 1. Show
that
1k + 2k + · · · + (p − 1)k
is a multiple of p.
Proof. Let S := [1]k + [2]k + · · · + [(p − 1)]k in Zp . The problem is equivalent to showing
that S = [0]. Let [a] be a generator of U (Zp ). Since [a] has order p − 1 and k < p − 1,
we know that [a]k is not the identity element [1]. Now multiply S through by [a]k and
call the result T . We find, by the distributive law, that T := [a]k +[2a]k +. . . [(p−1)a]k .
Notice now that since a is relatively prime to p, the list [a], [2a], [3a], . . . , [(p − 1)a] is
the same as the list [1], [2], . . . , [p − 1] of U (Zp ), in potentially a different order. It
follows S = T . Thus, in Zp , we have the equation
T − S = ([a]k − [1])S = [0].
We have already noted that [a]k is not [1]. Since Zp is a field, we must have S = [0],
which is what we wanted to prove.
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