Idea of Solution
(1) (Problem) Show that in a finite cyclic group of order n there is a unique
subgroup of order d for every divisor d of n.
Ans. Let H be any subgroup of G of order d. Let < a >= G. Then H =< ak >
n
. But since order of H is d. We get
for some k ∈ Z. Now, o(ak ) =
(k, n)
that (k, n) = n/d.
Consider K =< an/d >. Since n/d divides k, clearly H ⊆ K. But
number of elements in H and K are equal. Therefore, H = K.
(2) (Tutorial 1 Problem) Let n be a positive integer, and let a ∈ Z with
gcd (a, n) = 1. Prove that if k is the smallest positive integer for which
ak ≡ 1(mod n), then k/φ(n). (Here, φ is the Euler phi-function.)
Ans. To show that k/φ(n) one needs to show that total number of elements
of Z∗n is a multiple of k. So, first use the given hypothesis to show that
/ {1, a, · · · , ak−1 }, show
1, a, a2 , · · · ak−1 are in Z∗n . Then for any b ∈ Z∗n , b ∈
that {b, ba, · · · , bak−1 } is disjoint from {1, a · · · , ak−1 } and consists of distinct elements. The proof then follows easily.
(3) (Tutorial 4 Problem) Prove that Z∗2n (n ≥ 3) is not cyclic.
Ans.1 Proof by induction. Show that Z8 is not cyclic. Assume that Z2k is cyclic
for all k ≤ n. Then to show that Z2n is not cyclic.
Let Z2n be cyclic. Then there exists and element x ∈ Z such that
n−1
n−2
o(x) = O(Z2n ) = 2n−1 . This means 2n /(x2 − 1) = (x2 − 1)(x2 + 1).
n−2
Since Z2n−1 is not cyclic, we know 2n−1 6 |(x2 − 1). Therefore, at least
4/x2 + 1. This implies x2 = 4k − 1 = 4r + 3 for some r.
But square of an integer cannot be of the form 4r + 3. Therefore Z2n is
not cyclic.
Ans 2 Alternately, assume that Z∗2n is cyclic. But a finite cyclic group cannot
have two elements of order 2 (make sure you understand why?). But both
2n−1 − 1 and 2n−1 + 1 will have order 2.
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