MATH 366 Homework 10 Solutions Spring ’11
Section 6:
52) Let p be a prime number. Find the number of generators of the cyclic group Zpr , where r is an integer
with r ≥ 1.
Claim: Zpr has pr − pr−1 generators.
Proof: We will have as many generators of Zpr as elements that are relatively prime to pr (by Corollary
6.16). We claim that the only elements that are not relatively prime to pr are multiples of p. Clearly, a
multiple of p is not relatively prime to pr since their gcd must be at least p. Conversely, the divisors of pr
that are in Zpr are 1, p, p2 , . . . , pr−1 . Thus, for x ∈ Zpr , if gcd(p, x) 6= 1 for x ∈ Zpr then gcd(p, x) = ps for
some s ∈ {1, 2 . . . , r}, so ps |x so x = kps for some integer k, so x = (kps−1 )p. Since k, p ∈ Z, kps−1 ∈ Z.
Thus x is a multiple of p, proving that the only elements that are not relatively prime to pr are multiples
of p. The smallest multiple of p in Zpr is 0p = 0. The largest is pr − p = (pr−1 − 1)p since if we add one
more p to pr − p we’ll get pr − p + p = pr which is not an element of Zpr . Thus, the multiples of p in Zpr
are 0, p, 2p, . . . , (pr−1 − 1)p, and so there are pr−1 − 1 + 1 = pr−1 multiples of p in Zpr . This gives us pr−1
fewer generators of Zpr . Thus, the number of generators of Zpr is pr − pr−1 .
55) Show that Zp has no proper nontrivial subgroups if p is prime.
Proof: Since Zp is cyclic (Zp = h1i), all its subgroups must be cyclic. By Thm 6.14, for a 6= 0 ∈ Zp ,
p
|hai| = gcd(p,a)
since |Zp | = p and a = a(1), where 1 is a generator of Zp . But gcd(p, a) = 1 for all a ∈ Z
p
since p is prime, so |hai| = gcd(p,a)
= p1 = p for all a ∈ Zp . Thus, hai must be all of Zp for all a ∈ Zp , i.e.
Zp has no proper non-trivial subgroups.
Section 8:
16) Find the number of elements in the set {σ ∈ S4 : σ(3) = 3}.
If the element 3 is fixed, we are essentially just finding the number of permutations on 3 elements as
opposed to 4. Thus, there are 3! = 6 elements in the set {σ ∈ S4 : σ(3) = 3}.
36) Show by an example that every proper subgroup of a nonabelian group may be abelian.
Example: Consider D3 . It is not abelian since ρ1 µ1 = µ3 6= µ1 ρ1 , but its only proper subgroups are
{ρ0 } = hρ0 i, {ρ0 , ρ1 , ρ2 } = hρ1 i, {ρ0 , µ1 } = hµ1 i, {ρ0 , µ2 } = hµ2 i, {ρ0 , µ3 } = hµ3 i, which are all abelian since
they are all cyclic.