# Is this ok to prove?

1. Dec 4, 2004

### semidevil

so I need to show that in any group, it's elements and inverse has the same order.

so can I say that since it is a group, we know that there exists a unique inverse for each element. So each element would have 1 inverse. And then, that means we have the same number of elements as number of inverses?

does that work? or am I missing something?

2. Dec 4, 2004

### shmoe

Can you state the definition of the order of an element in a group?

3. Dec 4, 2004

### semidevil

so the order is the number of elements in a group....

is this suppose to be a hint? :)

4. Dec 4, 2004

### lokisapocalypse

semidevil...

The order of an element g in a group G is n such that g^n = identity where n is the smallest positive integer >= 1. To prove an element and its inverse have the same order you can say:

g^n = identity
(g^-1) ^ n = (g^n)^-1 = (identity)^-1 = identity

5. Dec 4, 2004

### shmoe

That's the order of the group, not an element. I was hoping to get you to check the definition carefully;).

6. Dec 4, 2004

### lokisapocalypse

Heh, sorry shmoe...didn't mean to steal your thunder.

7. Dec 4, 2004

### shmoe

No thunder to be stolen, we're all here to help (or be helped)