# Does a Finite Field of 36 elements exist? [duplicate]

I'm having a little trouble understanding the finite fields theory, so I'm sorry if my question would seem a little stupid.

I wanted to know if a finite field of 36 elements could exist. Basically, I thought of the field $$\mathbb{F}_{37}$$, which has 37 elements. But does for example the invertibles of $$\mathbb{F}_{37}$$ form a field of 36 elements or not? Since all elements are invertible and we have 36 invertibles in $$\mathbb{F}_{37}$$.

• The invertibles of $\mathbb{F}_{37}$ is a group under multiplication, with 36 elements. But it is do not form a field, because it does not contain the neutral for addition (0), wich is (thus) not even an internal law. The present observation does not answer the question, but this does. – fgrieu Feb 25 '19 at 11:31

Finite fields only exist for an order $$q$$ if $$q$$ is prime or a power of a prime (http://mathworld.wolfram.com/FiniteField.html). Since 36 is neither a prime itself, nor it is a power of a prime ($$36 = 2^3 * 3^2$$), there is no finite field with 36 elements.