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Abelian Groups Cyclic


Date: 03/05/2002 at 11:48:24
From: kristin bedell
Subject: Abelian Groups

Prove that every abelian group of order 6 is cyclic.

I know the non-abelian group of order 6, but I don't know how to find 
it.

Help please - I've been racking my brains trying to figure this out.

-Kristin


Date: 03/05/2002 at 18:13:55
From: Doctor Paul
Subject: Re: Abelian Groups

Suppose by way of contradiction that |G| = 6 and G is not cyclic.  
Then G has no elements of order six. The identity is the only element 
of order one. By Lagrange's Theorem, the order of each element (which 
is the order of the subgroup it generates) must divide the order of 
the group so if |G| = 6 G is not cyclic then g in G implies |g| = 2 
or |g| = 3.

Claim:

It cannot be the case that every non-identity element has order two.

Proof:

Assume that all non-identity elements have order two.

Pick two such elements h and k. So h and k are distinct and both have 
order two. Notice that h^2 = 1 ==> h = h^(-1). Similarly, k = k^(-1)

Let H = {1, h, k} be a subset (notice, not a subgroup since hk is not 
in H) of G. Notice that hk is not equal to h and hk is not equal to k:

If hk = h then k = 1 (multiply both sides by h^(-1) on the left) 
which is not the case, and if hk = k then h = 1 (multiply both sides 
by k^(-1) on the right), which is not the case either.

Also notice that hk is not 1, for if hk = 1 then h = k^(-1) = k, which 
is not true since we chose h and k to be distinct.

So hk is not 1, hk is not h, and hk is not k. Notice also that hk has 
order two:

hk is not 1 and (hk)^2 = h^2 * k^2 since G is abelian and this is 
clearly the identity. Thus hk has order two and is hence an element 
of G.

So now let H' = {1, h, k, hk} be a subset of G.  H' contains four 
distinct elements of G. But we know |G| = 6 so there exists m in G 
with m not in H' and |m| = 2 (G has six elements and we have only 
discussed four of them, so we must be able to pick another one).  
So m is not 1, m is not h, m is not k, and m is not hk.  

Also: mh and mk are going to be elements of G since G is closed under 
the group operation (mhk will also be in G, but we don't need this 
information). If we can show that mh and mk are distinct and not equal 
to m, and that they are not any of the elements in H', then we will 
have that 

G must contain *at least* seven elements:  {1, h, k, hk, m, mh, mk}

It may also be the case that mhk will be in G, but we would need to 
check to see if mhk is one of the other elements we already have.  
This isn't necessary to get the desired contradiction.

Clearly if G contains at least seven elements, then we have a 
contradiction since we assumed |G| = 6. So let's show that mh and mk 
are distinct and not equal to m, and that they are not any of the 
elements in H'.

First: mh and mk are distinct. For if mh = mk then h = k, a 
contradiction.

Certainly mh is not m, because if this were the case, then h would 
equal 1, which we know isn't true.  mh is not 1 because if mh = 1 
then we have m = h^(-1) = h, a contradiction.  mh is not h because if 
mh = h, then m = 1, a contradiction.  mh is not k, because if mh = k, 
then m = k*h^(-1) = k*h = h*k, a contradiction. Finally, mh is not hk 
because if mh = hk, then hmh = hhk, which implies m*h^2 = h^2*k, 
which implies m = k, a contradiction.

A similar argument shows that mk is not m, and that mk is not any of 
the elements in H'.

So if every non-identity element of G has order 2, then we can get 
that |G| > 6, a contradiction. So it must the case that G cannot have 
all non-identity elements of order two.

Now we want to argue that no non-identity element can have order 
three.

Once we have this, then it must be the case that G contains at least 
one element of order two and at least one element of order three 
(since we will have shown that G cannot have all non-identity elements 
of order two and G cannot have all non-identity elements of order 
three and the only possibilities for the orders of non-identity 
elements in G are two and three). Let x and y be in G and let |x| = 2 
and |y| = 3 (notice that this forces x^(-1) = x and y^(-1) = y^2).  
Then xy is in G (since G is a group) and we want to claim that 
|xy| = 6.  Notice that:

xy is not the identity, for if it were, then xy = 1, which implies 
x = y^(-1) = y^2. So we have x = y^2. Notice that |x| = 2 implies 
x = x^(-1). Substituting, we obtain:

x^(-1) = y^2, which implies x*y^2 = 1, and we'll see below that this 
cannot be.

(xy)^2 = x^2 * y^2 = y^2, which is not the identity; if y^2 = 1, then 
y = y^(-1), and we know that in this case y^(-1) = y^2, so we would 
have y = y^2, which would imply y = 1 (multiply both sides by y^(-1), 
which is plainly not true.

(xy)^3 = x^3 * y^3 = x, which is not the identity.  

(xy)^4 = x^4 * y^4 = y, which is not the identity.  

(xy)^5 = x^5 * y^5 = x*y^2, which is not the identity; if x*y^2 = 1, 
then we can write:

x = y^(-2) = [y^(-1)]^2 = [y^2]^2 = y^4 = y

Plainly, x is not y because |x|=2 and |y| = 3.

Finally, (xy)^6 = x^6 * y^6 = 1 * 1 = 1.

So |xy| = 6.

This is a contradiction since we assumed that G was not cyclic.

Thus our assumption that G was not cyclic must be false, and it is in 
fact the case that G is cyclic.

Now we come back to the unresolved issue above of showing that no non-
identity element can have order three.

Suppose by contradiction that every non-identity element has order 
three.

Let h be in G and let H = {1, h, h^2} be a subset of G. Then the 
elements in H are distinct. Moreover, 

|G| = 6, so there exists k in G, k is not in H, and |k| = 3.

Plainly, G contains k^2, and I claim that k^2 is not in H.  

If k^2 = 1, then k = k^(-1), which is not true since we know that 
k^(-1) = k^2 and k^2 is not k since |k| = 3.

If k^2 = h, then square both sides to obtain: k = h^2, a contradiction 
of the fact that k is not in H.

If k^2 = h^2, then square both sides to obtain: k = h, a contradiction 
of the fact that k is not in H.

Thus we have H' = {1, h, h^2, k, k^2} is a subset of G.

Again, |G| = 6 so there exists m in G, m is not in H', and |m| = 3.

Similar arguments show that m^2 is not 1, m^2 is not h, m^2 is not 
h^2, m^2 is not k, m^2 is not k^2.

Thus we have the subset of elements:

{1, h, h^2, k, k^2, m, m^2} all in G. This cannot be in a group of 
order six. Notice that we have yet to consider any of the products of 
the form k*h, k*h^2, etc. But we didn't need to in order to show that 
G must contain more than 6 elements.

Thus no non-identity element can have order three.

This completes the (somewhat lengthy) proof. I hope it's clear. Please 
write back if you'd like to talk about this some more. 

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Modern Algebra
College Number Theory

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