G is isomorphic to a field

Hello

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Consider G, a Domain which contains the one and a finite number of elements. Now Show that G is isomorphic to a field containing $\displaystyle q^n$ elements where q is a prime and n is a natural number.

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A domain firstly means that G is a ring containing no left- or right- zerodivisors but the 0 itself. So intuitively if there is a number m that divides q there exists an element which can be multiplicated with q and since the characteristic of $\displaystyle \mathbb{Z}_{q^n}$ is q that has to be zero?

Yet I don't see a way to Show this in an understandable, formal correct way. Could please someone give me a hint what to do in this case here?

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Re: G is isomorphic to a field

Hi,

I think the attachment solves your problem:

Attachment 29378