Z_2007 Subgroup of Order 6: True or False?

[raynard]

Homework Statement

True or false:
U($$\mathbb{Z}_{2007}$$), . (=group of units) has a subgroup of order 6.

Homework Equations

We know that the $$\phi(2007)$$ (= Euler's tolient function) = 1332, which is the amount of elements in U($$\mathbb{Z}_{2007}$$), .

The Attempt at a Solution

We clearly see that 6 could be a valid subgroup of U($$\mathbb{Z}_{2007}$$), . , as it is a divisor of the total amount of elements in that group. However, I fail to find any example of such a group (I can't find any example of an element a for which $$a^6 = 1$$, other than the trivial case).
Any ideas?

 

[micromass]

Maybe you should start by writing $$\mathbb{Z}_{2007}$$ under the form $$\mathbb{Z}_n\times \mathbb{Z}_m\times ...$$ where n,m,... are powers of primes.

The reason you do that, is because of the following fact:
The order of an element (a,b) in a group GxH, is the least common multiple of the order of a and the order of b...

 

[raynard]

The order of an element (a,b) in a group GxH, is the least common multiple of the order of a and the order of b...

Could you explain me why this is so? (We haven't seen any kind of such a theorem)

 

[Tedjn]

You should be able to convince yourself of this fact. First, show that (a,b) raised to the least common multiple power does equal the identity (1,1). Second, show by contradiction that anything less won't do.

I haven't touched algebra in about a year, so I can't help much more on this interesting approach. I will say that by trial and error you will find that 40 works as an example.

 

[raynard]

I still don't have a clue on how to get further in this problem.

This is what I've got so far:

assuming there exists a subgroup of order 6, all elements should have order 1, 2, 3 or 6. The identity element is the only one having order 1, so that leaves us with 5 other elements. Now there are several possibilities:

1. There is only one more element a with order 6. This would imply that $$a^6 = 1 + n*2007$$. I don't manage to find such an a.
2. All remaining elements have order 2. This would imply that there is an $$a^2 = 1 + n*2007 \Rightarrow 2007|(a-1)(a+1) \Rightarrow 3 | (a+1) \vee 223 | (a+1)$$. This way, we find that for instance 224 is an element of order 2. However, I fail to find any other elements, let alone that their product is still element of the subgroup.
3. ...

I assume my method is wrong, but I fail to see any other way to find a subgroup (or to prove there aren't any)

 

[micromass]

Didn't Tedjn already answer your question?? 40 is the element you're looking for...

Some advanced algebra guarantees that an element of order 6 exists, but it doesn't say how to find it. To find it, I see no other method than trial and error.

 

[raynard]

Could you explain how to do this 'trail and error' ?

 

[micromass]

Try every elementand see if they generate a subgroup of order 6...