Am I heading in the right direction:

If the order \(\displaystyle n\) of any element \(\displaystyle g\) of the group \(\displaystyle G_p\), where \(\displaystyle p\) is the order of the group \(\displaystyle G\), is such that \(\displaystyle g^n=E\), and by Legrange's theorem \(\displaystyle n\) must be divisible into \(\displaystyle p\), then for \(\displaystyle p\) prime, it must be true that \(\displaystyle n\) is either 1 or \(\displaystyle p\).

Since every element \(\displaystyle g_k\) of the Group \(\displaystyle G_p\) must have an order \(\displaystyle n\), and since \(\displaystyle n\) is restricted to either 1 or \(\displaystyle p\) when \(\displaystyle p\) is prime, all elements (except E) must have the same order -- i.e. \(\displaystyle p\).

Now, need to show that only one such Group \(\displaystyle G_p\) can satisfy this condition.

??