MHB Finite Fields - F_4 - Galois Field of Order 2^2

[Math Amateur]

I am reading Beachy and Blair's book: Abstract Algebra (3rd Edition) and am currently studying Section 6.5: Finite Fields,

I need help with a statement of Beachy & Blair in Example 6.5.2 on page 298.

Example 6.5.2 reads as follows:https://www.physicsforums.com/attachments/2858In the above example, Beachy and Blair write the following:

" ... ... Note that since the multiplicative group of non-zero elements has order 3, it is cyclic. ... ... "

Can someone explain why this assertion follows?

Peter

 

[Deveno]

It is elementary group theory.

Theorem:

If $G$ is a group of order $p$, then $G$ is cyclic.

Proof:

Let $x \neq e$ be any non-identity element of $G$.

By Lagrange's theorem, the order of $\langle x\rangle$, the cyclic subgroup generated by $x$, has order dividing the order of $G$.

Since this is prime, either $|x| = 1$, or $|x| = p$ (by definition, the order of $x$ as an element, is also the order of $\langle x\rangle$ as a subgroup*).

If the order of $x$ is 1, then $x^1 = x = e$, contradicting our choice of $x$. Hence $|x| = p$, so that $\langle x\rangle = G$, so that $G$ is cyclic (with generator $x$).

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Since 3 is prime, any group of order 3 is cyclic, QED.

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*Note: perhaps this is not obvious. Suppose that $k$ is the smallest positive integer for which we have: $x^k = e$.

Now if $\{e,x,x^2,\dots,x^{k-1}\}$ are not all distinct, we have: $x^m = x^n$ for some $0 \leq m < n \leq k-1$.

Then $0 < n-m < k-1$, and we have:

$x^{n-m} = x^n(x^{-m}) = (x^n)(x^m)^{-1} = (x^n)(x^n)^{-1} = e$, contradicting our choice of $k$.

On the other hand, suppose $\langle x\rangle = \{e,x,x^2,\dots,x^{k-1}\}$, and these are distinct.

Since a group is closed under multiplication, we have $x^k \in \langle x\rangle$.

If $x^k = x^m$, for $0 < m \leq k-1$, then:

$e = (x^k)(x^k)^{-1} = (x^k)(x^m)^{-1} = x^k(x^{-m}) = x^{k-m}$, where $0 < k-m \leq k-1$, contradicting distinctness.

Thus the only viable possibility is $m = 0$, that is $x^k = e$, and no smaller positive integer has this property.