Let a b Z Show that the subgroups a and b generated by a and

Let a, b Z. Show that the subgroups ([a]) and ([b]) generated by [a] and [b] in Z_n are equal, if and only if g.c.d.(a, n) = g.c.d.(b, n). Show that ([b]) = Z_n if and only if g.c.d.(b, n) = 1. This is the assertion of Corollary 2.2.28 (c).

Solution

Solution: Let a, b belongs to Z. Also let <[a]> and <[b]> be two subgroups of Z_n generated by [a] and [b] respectively. Also order of the group Z_n = n.

Suppose that <[a]> and <[b]> are equal. Then number of elements in <[a]> and <[b]> are equal say m.

then order of <[a]> and order of <[b]> devides the order of group Z_n. Since Zn is cyclic group of order n,

<[a]> and order of <[b]> must be cyclic.

Thn o([a]) = n/g.c.d(a,n) and o([b]) = n/g.c.d(b,n).

But o([a]) = o([b])

implies that   n/g.c.d(a,n) = n/g.c.d(b,n)

Hence g.c.d(a,n) = gcd(b,n).

Conversely suppose that g.c.d(a,n) = gcd(b,n) = m(say).

Then m devides n and a, b.

Also subgroups generated by [a] and [b] are equal.

then order of [a] and [b] devides

Second part: Suppose that <[b]> = Zn. Then [b] is a genarator then o([b]) = o(Zn) =n.

But o([b]) = n/gcd(b,n) = n.

Hence gcd(b,n) =1

Conversely suppose that gcd(b,n) =1.

o([b]) = n/gcd(b,n) = n/1=n

Then [b] must be a generator of Zn.

So <[b]> = Zn