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Cyclic Groups

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 

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 

  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 
Associated Topics:
College Modern Algebra

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