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). Let b e Z, b notequalto 0. The cyclic subgroup ([6]) of Zn generated by [6] is equal to the cyclic subgroup generated by [d], where d = g.c.d. (b, n). The order of [b] in Z_n is n/g.c.d.(6, n). In particular, ([b]) = Z_n if and only if b is relatively prime to n.

Solution

Given that gcd of (a,n) = gcd of (b,n)=d (say)

Since gcd (a,n) = d , group generated by a will have order d.

Similarly gcd (b,n) = d group generated b also will have order d.

Groups generated by a and b have the same order.

This implies a^d = b^d =e
So a , a^2, a^3.... = b, b^2, b^3 ... though may not be in the same order.

Hence <a> = <b>

------------------------------

If gcd (b,n) = 1

then <b> is the cyclic group generated by 1

Or <b> = zn