Does Group Cardinality Determine Element Order?

[RJLiberator]

Homework Statement

If G is a group with n elements and g ∈ G, show that g^n = e, where e is the identity element.

Homework Equations

The Attempt at a Solution

I feel like there is missing information, but that cannot be.
This seems too simple:
The order of G is the smallest possible integer n such that g^n = e. If no such n exists, then G is of infinite order.

From this definition of order can we simply state that since G is a group with 'n' elements then there must exist an n such that g^n = e ?

order is denoted as °(g)
So
°(g) = n ==> g^n = e.

 

[vela]

Homework Statement

If G is a group with n elements and g ∈ G, show that g^n = e, where e is the identity element.

Homework Equations

The Attempt at a Solution

I feel like there is missing information, but that cannot be.
This seems too simple:
The order of G is the smallest possible integer n such that g^n = e. If no such n exists, then G is of infinite order.

Aren't you mixing up G and g here? You're describing the definition of the order of the element g, not the order of the group G.

From this definition of order can we simply state that since G is a group with 'n' elements then there must exist an n such that g^n = e ?

order is denoted as °(g)
So
°(g) = n ==> g^n = e.

 

[HallsofIvy]

Where did you get this: "The order of G is the smallest possible integer n such that g^n = e"? The definition of "order of a group" is simply the number of elements in the group (the cardinality of the underlying set). It looks to me like you are being asked to prove the statement you give.