# Throwing out an element of a field

I was reading my book on Elementary Algebra and saw this theorem:

Suppose that $$F$$ is a finite field of order $$q$$, then the group $$F^*$$ is a cyclic group of order $$q-1$$.

I don't understand the transition from a field to a group. How come the theorem says $$F^*$$ is a group when the element $$0$$ is omitted?

• Please clarify: are you asking why it is a (cyclic) group, or why $0$ had to be omitted? – Bill Dubuque Jan 22 at 16:17

## 6 Answers

Well, $$\mathbb{F}^*$$ is the group $$(\mathbb{F}\setminus \{0\},\cdot)$$ hence you are only considering multiplication as operation, in particular, you need to throw out $$0$$. In more general algebra (Rings) this still holds, but there you throw out every element which is not invertible. hence you can imagine $$\mathbb{F}^*$$ as the multiplicative group of invertible elements.

$$F^\ast$$ is a set with a binary operation (the multiplication operation on $$F$$) which is associative (since it's associative on all of $$F$$) and in which all elements have inverses.

That's a group.

It wasn't a group before when you had all of $$F$$ since $$0$$ did not have an inverse. $$F$$ was a monoid though, with its multiplication operation.

If $$F^{*}$$ were to be a field, it would need an additive identity. But this would have to be the additive identity of the original field $$F$$, which as you pointed out, is omitted from $$F^{*}$$ since it isn't a unit. So $$F^{*}$$ is just a group.

Every field is a group if you remove the $$0$$ element and restrict it to just the multiplication operation because:

The definition of a group is:

A set of elements $$G$$ and an associated binary operation, $$*$$, so that:

1) $$*$$ is associative.

2) There is an element $$e\in G$$ so that for any $$g\in G$$ we have $$e*g=g*e = g$$.

3) For every $$g \in G$$ there is an element $$g'$$ so that $$g*g' = g'*g = e$$.

The definition of a field is

A set of elements $$F$$ and two associated binary operations, $$+, \cdot$$ so that

Addition axioms.

A1) $$+$$ is associative.

A2) $$+$$ is commutative.

A3) There is an element, $$0 \in F$$ so that for all $$q\in F$$ that $$q+0=q$$.

A4) For every $$q \in F$$ there is an element $$-q\in F$$ so the $$q+ (-q) = 0$$.

[Notice: A1, A3, A4, are the same axioms (with different notation) as the axioms for a group. Thus we can say a Field $$F$$ with only the addition operation considered forms a Group (with an addition condition that it is commutative).]

Multiplication Axioms

M1: $$\cdot$$ is associative.

M2: $$\cdot$$ is commutative.

M3: There exists an element $$1 \in F$$ so that for all $$q \in F$$, $$1\cdot q = q$$.

[Some text explicitly state as an axiom that $$1 \ne 0$$. Other texts will prove that if $$1=0$$ then then Field only has one element and is therefore trivial and not worth considering (The field would just be the set $$\{a\}$$ with the properties $$a+a=a$$ and $$a\cdot a = a$$ and there is absolutely nothing more to say about it). We'll take it for granted that $$1 \ne 0$$]

M4: For every element $$q\in F$$ so that $$q\ne 0$$ there is an element $$q^{-1} \in F$$ so that $$q*q^{-1} = 1$$.

[The axiom doesn't state that $$q^{-1} \ne 0$$. However it can be proven easily that $$q*0 = 0 \ne 1$$ so we can assume that \$q^{-1} \ne 0.]

Note: M1, M3, M4 when applied to the set $$F\setminus \{0\}$$ are the same axioms as the axioms of a group with only different notation. So we can state that $$F^*$$, the set $$F\setminus \{0\}$$ associated with the binary operation $$\cdot$$, is a group (with the additional condition $$\cdot$$ is commutative).

The final axiom:

D: For all, $$q,r,s \in F$$, $$q\cdot (r + s) = (q\cdot r) + (q\cdot s)$$.

......

What the theorem is saying is that if $$F$$ is finite then the group is cyclic. Which is not so trivial.

Once you remove the $$0$$ of the field, you have a group under multiplication (the same multiplication as defined on the original field, that is), assuming your field is not the trivial field (i.e. a field with $$0$$ as its only element).

This is because, in a field, every non-zero element is invertible, so you can reinterpret $$F\backslash\{0\}$$ in a more useful light- as being the set of all invertible elements in $$F$$. If $$a$$ is invertible and $$b$$ is invertible, $$ab$$ is also invertible, as $$(ab)^{-1}=b^{-1}a^{-1}$$- this means that $$F$$ is closed with respect to products. Also, if $$a$$ is invertible, so is $$a^{-1}$$ (since $$((a^{-1})^{-1}=a$$). Also, multiplication on any field is defined to be associative so it remains associative in $$F\backslash\{0\}$$. Finally, since $$F$$ is not the trivial field, $$F$$ has at least one non-zero element, so $$F\backslash\{0\}$$ is non-empty. Suppose it has some element, $$x$$, in it. By the arguments already outlined, it must be the case that $$x^{-1}\in F$$ and, then, that $$x*x^{-1}=1\in F$$, so $$F\backslash\{0\}$$ has a neutral element (again, as the multiplication on $$F\backslash\{0\}$$ is the same on that on $$F$$, $$1$$ is the neutral element of $$F\backslash\{0\}$$).

(the assumption that $$F$$ is finite will come in handy in the next part of the argument)

So $$F\backslash\{0\}$$ is a group. But why a cyclic one?

You can actually make a more general claim than the one for which you ask for a proof in your question- that is, that any subset of a finite field, $$F$$, that acts as a group under multiplication (like $$F\backslash\{0\}$$) is in fact a cyclic group under multiplication (but this will require a smidge of information from the theory of rings of polynomials).

The proof (at least, the one that I've seen) is as follows. Suppose we have a subset, $$G$$, of a field, $$F$$, that is a group under multiplication. Since $$F$$ is finite, this group will also be finite. Let its number of elements be denoted by $$n$$ (some finite number).

Notice that the polynomial equation $$x^k-1=0$$ (as an element of $$F[X]$$) can have no more than $$k$$ distinct solutions as its degree is $$k$$ (this was the thing that needed a smidge of polynomial ring knowledge).

We also know that any solution, $$y$$, to this equation will have a multiplicative order that divides $$k$$, as $$y^k-1=0\implies y^k=1$$ (and, as mentioned before, $$1$$ must be the neutral element of any group under multiplication that is a subset of $$F$$).

But, note that if $$k\not | n$$, then, $$x^k-1=0$$ will have no solutions in $$G$$ as every element in $$G$$ must have an order that divides $$n$$ (as the order of any element will equal the order of its corresponding cyclic group, which, given its status as a subgroup of $$G$$, must, by Lagrange's theorem must have an order dividing $$G$$'s).

So, suppose for some divisor, $$d$$, of $$n$$, we have at least one element, $$a$$, in $$G$$ with order $$d$$. Then, we must have at least $$d$$ distinct elements in $$G$$ with an order dividing $$d$$ as the list of elements $$(1,a,a^2,a^3,...,a^{d-1})$$ is a list of distinct elements (given $$a$$'s order) and each element's order divides $$d$$ as, for any $$r$$, $$(a^r)^d=a^{rd}=a^{dr}=(a^d)^r=1^r=1.$$ But, on the other hand,(given what was said above about equations like $$x^d-1=0$$), there can be no more than $$d$$ elements in $$G$$ whose orders divide $$d$$. So, if there is at least one element in $$G$$ of order $$d$$ (where $$d|n$$), there are exactly $$d$$. Otherwise, there are $$0$$.

What remains now is simple- note that, if $$\mathbb Z_{}$$ is the cyclic subgroup of $$\mathbb Z_n$$ consisting of all elements with additive order dividing $$d$$, then, treating $$\mathbb Z_n$$ as a group under addition (talking about the same $$a$$ as in the last paragraph), $$\cong \mathbb Z_{}$$ (forgive the loose notation). This means that the number of elements in $$$$ with order equal to exactly $$d$$ (and note that every element in $$G$$ with order exactly $$d$$ must be in $$$$) is equal to the number of elements in $$\mathbb Z_{}$$ with order exactly $$d$$.

(in case you're wondering, $$\mathbb Z_{}$$ looks like $$\{0,\frac{n}{d},\frac{2n}{d},...,\frac{(d-1)n}{d}\}$$)

So, denoting the number of elements in $$\mathbb Z_n$$ with order exactly $$d$$ (where $$d|n$$) with $$\mathbb Z(d)$$ and the number of elements in $$G$$ with order exactly $$d$$ with $$G(d)$$, we have either $$\mathbb Z(d)=G(d)$$ or $$G(d)=0$$, or, put another way $$G(d)\leq \mathbb Z(d)$$ always holds.

Now, what we argue is that we can partition $$G$$ into equivalence classes of elements based on their order. We can do the same with $$\mathbb Z_n$$. In each equivalence class in $$G$$ consisting of all elements of order $$d$$, we have $$G(d)$$ elements so the total number of elements across all equivalence classes must be $$\sum_{d|n}G(d)$$. Similarly, in $$\mathbb Z_n$$, the total number of elements in all equivalence classes must be $$\sum_{d|n}\mathbb Z(d)$$. Using the previous paragraph, then $$\sum_{d|n}G(d)\leq \sum_{d|n}\mathbb Z(d)$$

But, recall that these equivalence classes are disjoint sets whose union add up to $$G$$ or $$\mathbb Z_n$$ respectively so the total number of elements in either will be $$\Big|G\Big|=\Big|\mathbb Z_n\Big|=n$$.

So if for a single $$d$$, we have $$G(d)<\mathbb Z(d)$$, then we'll have $$\sum_{d|n}G(d)< \sum_{d|n}\mathbb Z(d)\\\implies n

A contradiction. So $$G(d)=\mathbb Z(d)$$ for all divisor of $$n$$ (even, non-divisors technically). In particular, $$G(n)=\mathbb Z(n)$$, i.e., there are as many elements in $$G$$ with order $$n$$ as there are in $$\mathbb Z_n$$- but there is always at least one element of additive order $$n$$ in $$\mathbb Z_n$$, i.e., $$1$$, so there is at least one element of order $$n$$ in $$G$$, say, $$b$$. Since $$\subset G$$, but also $$\operatorname{ord}(b)=n=\Big|G\Big|$$, we must have $$=G$$ and, so, $$G$$ is cyclic.

(please edit in or comment for any mistakes/ corrections)

By definition, a field is a commutative ring in which every nonzero element is a unit.

Thus if you consider the set of nonzero elements,it is a group with the multiplication operation as operation.

It is cyclic, as is the multiplicative group of any finite field.