Date: 06/25/2003 at 12:15:29 From: Ashley Subject: Cyclic Groups I am supposed to prove that every subgroup of a cyclic group is characteristic. Can you help?
Date: 06/26/2003 at 02:40:01 From: Doctor Jacques Subject: Re: Cyclic Groups Hi Ashley, The key fact is that we know that every subgroup of a cyclic group is itself cyclic (if you don't know this, please write back). Let G be a cyclic group with generator a - we write that as G = <a>. G may be finite, like Z_n, or infinite, like Z. Let H be a subgroup of G. As H is cyclic, it is generated by an element b, H = <b>. As b belongs to G, we can write b as some power of a: b = a^k Now, let f be an automorphism of G. We want to show that f(H) = H. As usual, we do this by showing that either subgroup is contained in the other. f(a) is an element of G, and therefore can be written as: f(a) = a^m for some integer m. The image of any element x = a^s of G is therefore: f(x) = f(a^s) = (f(a)^s) = a^(ms) = a^(sm) = x^m In particular, f(b) = b^m. As H is the set of powers of b, f(H) is the set of powers of b^m, i.e. f(H) = <b^m> As b^m belongs to H, f(H) = <b^m> is a subgroup of H. Now, as f is an automorphism, there is an automorphism g such that (g o f) is the identity automorphism. As before, we can write: g(a) = a^n for some integer n, and, in general, g(x) = x^n for all x in G. As g is the inverse of f, we have: (g o f)(x) = x^(mn) = x for all x in G. Can you use this to show that H is a subgroup of f(H), and therefore that H = f(H) ? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/
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