Cyclic Group Generators and negative exponents.

The set

G = {1,2,3,4,5,7,8}

(g, X9)

a generator

<2> = {1,2,4,8,7,5}

ok

<5> = {1,5,7,8,4,2}

ok

<2> = <2^-1> = <5>

This is where I no longer understand. The idea is that we don't need to calculate <5> as we know that <2> = <2^-1> = <5>, but I don't see why. In the text this is explained using a circle and where 1 would be at 12 o'clock, 2 would be at one move anti-clockwise around the circle and so 2^-1 would be one move clockwise around the circle. I understand that the inverse of 2 is 5 modulo 9, but I can't seem to make this analogy work.

Re: Cyclic Group Generators and negative exponents.

<a>=<a^-1> in all cases.

Why ? Well <a^-1> contains a (the group generated by an element contains the inverse of this element). So we have a=(a^-1)^k for some k in N (k less or equal to the cardinal of the group). So a power of a is also a power of a^-1, because a^n=(a^-1)^kn.

And vice versa for the other inclusion.