# Homework Help: Unable to find generators of {G, Xn}

1. Apr 18, 2009

### soopo

1. The problem statement, all variables and given/known data
Consider the group {G, Xn} where
G is the set containing the n - 1 residue
classes modulo n excluding 0.

Which members are generators of {G, Xn} when
n = 7?

3. The attempt at a solution

The right answers are 3 and 5.

I get 2 only.

My table of results

$$1: 1^1 = 1$$
$$2: 2^1 = 2, 2^2 = 4, 2^3 = 4, 2^4 = 2$$
$$3: 3^1 =3, 3^2 = 2, 3^3 = 4$$
$$4: 4^1 =4, 4^2 = 2$$
$$5: 5^1 = 5, 5^2 =4$$
$$6: 6^1 = 6, 6^2 = 1$$
$$7: 7^1 = 7, 7^2 = 7$$

2. Apr 18, 2009

### Billy Bob

$$2^3\neq4$$

$$3^3\neq4$$

Find $$3^4, 3^5,\dots$$ and $$5^3,5^4, 5^5,\dots$$

(And 7=0.)

3. Apr 18, 2009

### soopo

I do not mean with 3^3 3 to the power of 3 only.
I mean with it 3^3 3 to the power of 3 modulo 7.

4. Apr 18, 2009

### soopo

I cannot find $$3^4$$.
There is no such number which is 3 times 3 times 3 times 3.
The only way to get that is when n = 9, that is for 9 times 9.

5. Apr 18, 2009

### matt grime

You do understand the operation is multiplication and reduction modulo 7. It doesn't make sense to say that 'there is no such number in the set that is 3^4 mod 7'. There is. 3^2 is 2 mod 7. 3^3 is 6 mod 7, and 3^4 is 4 mod 7.

6. Apr 18, 2009

### soopo

I think I got it.

The order of a group is the number of order of elements.
For example, in this case,

2^1 = 1, 2^2 = 4, 2^3 = 1
However, the group already has the order 1 so we stop.

I get that 3 and 5 have 7 different results from modulo 7.
This means that the Cayley table is really useful here too, since I can use it repeatedly in the calculations.