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

In summary, Beachy and Blair assert that the multiplicative group of non-zero elements in Example 6.5.2 is cyclic because its order is 3, which is a prime number. This follows from the theorem in elementary group theory that states that any group of order p (where p is prime) is cyclic. The proof of this theorem is based on the fact that the order of an element is equal to the order of the subgroup it generates. Therefore, since 3 is prime, any group of order 3 is cyclic.
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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
 
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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 $x$).

*****************

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

*****************

*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.
 
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FAQ: Finite Fields - F_4 - Galois Field of Order 2^2

What is a finite field?

A finite field is a mathematical structure that consists of a finite set of elements and two operations, addition and multiplication, that follow certain rules. These fields are commonly used in cryptography, coding theory, and other areas of mathematics.

What is F4?

F4 is a finite field of order 2^2, also known as a Galois field. It contains four elements - 0, 1, α, and α^2 - where α is a root of the irreducible polynomial x^2 + x + 1.

How are finite fields used in cryptography?

Finite fields are used in cryptography to perform operations on large numbers in a secure and efficient manner. They are especially important in public key cryptography, where the security of the system relies on the difficulty of solving certain mathematical problems in finite fields.

Can you give an example of a real-world application of F4?

One example of a real-world application of F4 is in the design of error-correcting codes for communication systems. These codes rely on the properties of finite fields to detect and correct errors in transmitted data.

How do you perform addition and multiplication in F4?

In F4, addition and multiplication are performed using standard arithmetic rules, but with the additional property that all operations are performed modulo 2. For example, 1 + α = α, since 1 + α = 2, and 2 ≡ 0 (mod 2). Similarly, α^2 + α = 0, since α^2 + α = 1, and 1 ≡ 1 (mod 2). Multiplication is performed in a similar manner, but with the additional rule that α^2 = α + 1.

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