# Unable to find generators of {G, Xn}

## Homework Statement

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?

## The Attempt at a Solution

The right answers are 3 and 5.

I get 2 only.

My table of results

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

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$$2^3\neq4$$

$$3^3\neq4$$

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

(And 7=0.)

$$2^3\neq4$$

$$3^3\neq4$$
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.

$$Find [tex]3^4, 3^5,\dots$$ and $$5^3,5^4, 5^5,\dots$$

(And 7=0.)
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.

matt grime
Homework Helper
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.

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.
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.