# So exactly what am I doing here?

1. Dec 2, 2004

### semidevil

to show that U(14) = <3> = < 5 >.

i.e, U(14) is cyclic.

ok, so U(14) = 1, 3, 5, 9, 11, 13

and <3> = 3^1 = 3, 3^2 = 9, 3 ^3 = 13, 3^4 = 11, 3^5 = 5, and 3^6 = 1.

So what am I doing to show that it is Cyclic? Am I showing that with <3>, I can generate the same elements over and over? is that what it means that it is cyclic?

2. Dec 2, 2004

### matt grime

Surely your notes define cyclic?

A finite group is cyclic if there is an element whose order is the same as the order of the group.

Or a group is cyclic if there is a single element x such that {x^r | 1<=r<=ord(x)}=G. Ie all elements are a power of x for some x.

3. Dec 2, 2004

### mathwonk

i.e. if the map taking n to r^n is surjective, then your group is cyclic with generator r.

4. Dec 5, 2004

### Palindrom

I'm sorry, but I can't see the problem. You've shown that U(14)=<3>. You can show similarly that <3>=<5> in U(14).

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