1) let <a> be a cyclic group of order n. If n and m are relatively prime, then the function f(x) = x^m is an automorphism of <a>.
2) Let G be a group and a, b be in G. Let a be in <b>. Then <a> = <b> iff a and b have the same order.
The Attempt at a Solution
1) It is trivial to check that for any x, y in <a>, f(xy) = f(x)f(y). For injectivity, let x^m = y^m. We have x = a^k and y = a^j, for some integers k and j. Now a^(km) = a^(jm) implies that km = jm, which implies k = j, and hence x = y. (Since, all the powers of a are distinct, so if km and jm were distinct, then a^(km) = a^(jm) couldn't possibly hold).
I have a bit of a problem with surjectivity. When I look at examples of cyclic groups, the whole problem is very obvious intuitively - for any element, we can find adjust another one which will give the first one after multiplying m times - but, formally... Let y be in <a>, so y = a^k, for some integer k. We need to show that there exists some element a^j in <a> such that a^(jm) = a^k. I guess here the fact that m and n are relatively prime comes in... But I can't find the solution.
2) Since a is in <b>, a = b^k, for some k. Let <a> = <b>. Then, by definiton, the orders of these groups are equal, and by definition again, the orders of a and b must be equal. For the converse, let ord(a) = ord(b) = p. Is x is in <a> it is clearly in <b>, since a = b^k. Let x be in <b>. so x = b^j, for some j. Now, since the orders of b and a = b^k are equal, this means that there are p distinct powers of b and of b^k. But a power of b^k is a power of b, too, and hence <a> must equal <b>.