Prove G is a Cyclic Group

Date: 02/27/2003 at 17:11:41
From: Nicholas
Subject: Abstract Algebra/Group Theory

Let group G be finite Abelian such that G has the property that for 
each positive integer n the set {x in G such that x^n = identity} has 
at most n elements. Prove G is a cyclic group.

Date: 03/01/2003 at 08:03:25
From: Doctor Jacques
Subject: Re: Abstract Algebra/Group Theory

Hi Nicholas,

In any finite group (say of n elements), every element x of the group 
verifies x^n = e (this is a consequence of Lagrange's theorem).

A cyclic group is a group generated by a single element (for example 
a). It is the set of powers of a:

 {e, a, a^2, ......}

along with their inverses. We use the notation <a> for such a group.

If a cyclic group is finite, there is a smallest integer n such that 
a^n = e, and the group consists of the elements:

  <a> = {e, a, a^2, ..., a^(n-1)}

These n elements are all different, and the size of the group is 
therefore n. As

  a^i*a^(n-i) = a^n = e

we see that inverses are automatically included.

The order of an element is the order of the cyclic subgroup it 
generates (in the above example, the order of a is n).

We also know that all elements x of <a> satisfy the equation x^n = e.

Sorry to repeat that if you already know it, I just wanted to be 
sure.

Let us go back to the problem at hand.

What we are given is the fact that, for every n, the equation x^n = e 
has at most n solutions.

As G is finite, we can find at least one element of maximal order. Let 
a be such an element, <a> the subgroup it generates, and its order be 
n.

If <a> = G, G is cyclic by definition and we are done.

Assume, for the sake of contradiction, that <a> is not equal to G.

We can therefore find an element b not in <a>. Let b be of order m.

As we chose n to be the maximal order of elements of G, we have 
m <= n.

It is known that, in an Abelian group, if two elements have orders m 
and n, there exists an element of order equal to:

  p = lcm(m,n).

(If you are not familiar with that property, please write back, and 
I'll explain it).

In this case, as n is the maximal order of an element, we must have
p = n (obviously, lcm(m,n) >= n).

The lcm of m and n will be equal to n if and only if m divides n.

This means that n = km for some k, and:

  b^n = b^(km) = (b^m)^k = e^k = e.

and this shows that b is another solution of the equation x^n = e.

As all the n elements of <a> already satisfy that equation, and b is 
not among them, the equation has at least n+1 solutions, contrary to 
the hypothesis.

Does this help?  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/ 

Date: 06/21/2003 at 03:00:15
From: William Cheung
Subject: Modern Algebra - Group Theory

Dear Dr. Jacques and associates, fellow mathematicians,

You wrote in a reply to a group theory problem:

It is known that, in an Abelian group, if two elements have orders m 
and n, there exists an element of order equal to:

  p = lcm(m,n).

I managed to solve the original problem, except that I couldn't prove 
the omitted property related to the existence of the "lcm element."

I had the gut feeling that "lcm-order element" must exist in abelian 
group, otherwise the original theorem would be wrong.  I can see that 
there are two approaches to prove its existence:

1) constructive approach: for any two elements a, b in Abelian Group 
G of order m and n, respectively, provide the closed form formula for 
the "lcm-order element"

I tried for many weeks to arrive at such a formula, in vain. But my 
best guess for the formula is, assume that gcd(m,n) = k, then gcd 
(m/k, n/k) = 1, therefore there exist integer i and j such that 
i * m + j * n = 1; the "lcm-order element" *may* probably be a^j * b^i 
(usually multiplicative notation). I have a problem proving that it 
is.

2) deductive approach: to conclude that it exists with some elegant 
group-theoretic arguments.

Thank you for your kind reply.

Date: 06/21/2003 at 08:14:48
From: Doctor Jacques
Subject: Re: Modern Algebra - Group Theory

Hi William,

Let us assume that a has order m and b has order n. We distinguish 
two cases:

Case 1 : gcd(m,n) = 1
---------------------
In this case, lcm(m,n) = mn.

Consider the intersection H of <a> and <b>. H is a subgroup, and we 
claim that H = {e}.

If x is an element of H, <x> is a subgroup of <a> and <b>, and so the 
order of x must divide m and n, by Lagrange's theorem.

As gcd(m,n) = 1, x has order 1, i.e. x = e.

Let now x = a*b^(-1), and let k be the order of x.. As the group is 
Abelian, x^(mn) = (a^m)*b^(-n) = e, and k divides mn.

We can write:

  y^k = (a^k) * b^(-k) = e
  a^k = b^(-k) = c

As c belongs to <a> and <b>, it belongs to H, and c = e. This means 
that a^k = b^k = e, and k is a multiple of m and n, and therefore a 
multiple of lcm(m,n) = mn. As we saw that k divides mn, this shows 
that k = mn, and x is the required element.

Case 2 : gcd(m,n) = d > 1
-------------------------
We will factor m and n as m = m1*m2 and n = n1*n2.

Let p be a prime factor of mn (i.e. a prime factor of m or n), and 
assume that p has exponent k1 in m and exponent k2 in n (one of these 
exponents may be 0).

If k1 >= k2, include p^k1 in m1 and p^k2 in n2; otherwise, include 
p^k1 in m2 and p^k2 in n1.

We can check the following:

* Every prime factor of mn appears in either m1 or n1, but not both,
  which means that gcd(m1,n1) = 1.

* m1 and n1 contain all prime factors of mn with their greatest
  exponent, and this shows that m1*n1 = lcm(m,n).

* Every prime power of m appears in either m1 or m2, but not both,
  therefore m = m1*m2, and, in the same way, n = n1*n2.

For example, if m = 12 = 2^2*3 and n = 18 = 2*3^2, we will have

  m1 = 2^2, m2 = 3
  n1 = 3^2, n2 = 2

All this allows us to conclude that a^m2 and b^n2 have orders m1 and 
n1, respectively, with gcd(m1,n1) = 1 and lcm(m1,n1) = lcm(m,n).

Can you continue from here? 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/