# MATH 3020 A: Homework 4 Soution Ideas

advertisement ```MATH 3020 A: Homework 4
Soution Ideas
Herstein, in Topics in Algebra poses the following exercise.
Let G be a finite abelian group in which the number of solutions in G of the equation
xk = e is at most k for every positive integer k. Prove that G must be a cyclic group.
The following steps provide a proof. Give details and reasons as indicated below. The source is
Jean-Pierre Serre, Cours d’Arithmetic.
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1. (Proof that d|n φ(d) = n) Let C be a cyclic group of order n and let d | n. C has a unique
subgroup Cd of order d.
Let Φd be the set of generators of Cd . Explain why
X
X
n = |C| =
Card(Φd ) =
φ(d) .
d|n
d|n
Hint: Every element of C generates a subgroup Cd for some d.
SOLUTION IDEA
Every element of C generates a cyclic subgroup of C of order some d, with d | n. Then
{Φd }d | n is a partition of C and
n = |C| =
X
Card(Φd ) .
d|n
As for each d | n there is precisely one cyclic subgroup of order d, Card(Φd ) = φ(d) and the
result follows.
2. (Consideration of the problem from Herstein) Let G be finite of order n satisfying the property
that for every positive integer k the set of elements x with xk = e has at most k elements.
(a) Let d be a divisor of n. If there exists an element x ∈ G of order d, consider the cyclic
group &lt; x &gt;= {e, x, x2 , . . . , xd−1 }.
Explain why every element of G of order d must belong to &lt; x &gt;.
SOLUTION IDEA
Let x ∈ G be of order d. Every element y ∈ &lt; x &gt; satisfies y d = e. Since &lt; x &gt;
has d elements, by the condition on G, if z ∈ G, z d = e then z ∈ &lt; x &gt;.
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(b) If there exists x in G of order d, then the number of elements in G of order d is φ(d).
Justify. For any particular d, the number of elements in G of order d is either 0 or φ(d).
SOLUTION IDEA
Let x ∈ G be of order d. Then y ∈ G is of order d if and only if y is a generator
of &lt; x &gt; as follows.
If y is of order d then y ∈&lt; x &gt; by (a). Since its order is d it is a generator of &lt; x &gt;.
Conversely, if y is a generator of &lt; x &gt; it has the same order as x, i.e., d.
The number of generators of &lt; x &gt; is φ(d).
(c) Let Ψd be the number of elements in G of order d. Explain why for all d, Ψd ≤ φ(d).
SOLUTION IDEA
If there are any elements in G of order d, then there are φ(d) of them by (b). Then
Ψd = 0 or Ψd = φ(d) from which for all d, Ψd ≤ φ(d).
X
(d) Explain why
Ψd = n You may use without proof that the order of an element of a
d|n
finite group divides the order of the group.
SOLUTION IDEA
The condition of having the same order defines an equivalence relation on theX
set of
elements in G. Consideration of the partition defined by this relation gives n =
Ψd .
X
But Ψd = 0 if d 6 | n, hence n =
Ψd .
d|n
(e) Explain why Ψd = φ(d) for all d | n.
SOLUTION IDEA
We have
n=
X
Ψd ≤
d|n
X
φ(d) = n, which implies
d|n
X
d|n
Ψd =
X
φ(d) .
d|n
As in general, Ψd ≤ φ(d), the equality holds only if Ψd = φ(d) for all d | n.
(f) What is the value of Ψn ? How does this relate to Herstein’s exercise? Was the condition
that G be abelian used anywhere?
SOLUTION IDEA
As Ψn = φ(n) 6= 0, G has an element of order n, i.e., G is cyclic. This gives a proof of
the exercise. The proof did not require that we assume G to be abelian.
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3. Let G be a finite subgroup of the multiplicative group C∗ of nonzero complex numbers. Prove
that G is cyclic.
SOLUTION IDEA
In C ∗ , the identity e is the number 1. The condition xn = e translates to xn = 1, i.e.,
2πk
x is an nth root of unity. There are n such roots, in particular the numbers e n i , for k an
integer 0 ≤ k ≤ n − 1. For any k, G has at most k solutions to xk = e. The result follows.
Alternatively, as every element of G is of finite order, it must be a root of unity. Let x
be minimal positive with eix ∈ G. It is possible to prove directly that eix is a generator of G.
Note: You may discuss these questions but must not share solutions. Do your own
writing. Work that shows evidence of having been copied will receive 0 points.
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