Verified answer

Where to start...

The symbol "<a>" means that the group is generated by powers of a.

In other words, <a> = {..., a^(-2), a^(-1), a^0, a^1, a^2, ...}.

This is an instance of an infinite cyclic group; all elements can be obtained by powers of one element (called "a" in our case) and there are infinitely many elements in this set.

This problem is asking to find all elements in this set whose powers will create every element in G = <a>.

There are two obvious answers to this: a itself and a^(-1).

Note that positive powers of a^(-1) will yield a^(-1), a^(-2), ... while negative powers of a^(-1) will give a^1, a^2, .... And, [a^(-1)]^0 = a^0.

It turns out that these are the only solutions to your question. Here's why this is true:

Suppose a^n is a generator for G. Then, there exist some integer power k such that (a^n)^k = a^(nk) = a.

==> a^(nk - 1) = 0 by cancellation laws.

==> nk - 1 = 0, since only the identity a^0 = e has finite order in an infinite cyclic group.

==> nk = 1.

The only way two integers multiply to give 1 if if n = k = 1or n = k = -1.

At any rate, we conclude that a = a^1 and a^(-1) are the only possible generators of G. Since this point was established earlier, we are donw with this problem.

I hope this is useful!