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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?

Well, $mathbbF^*$ is the group $(mathbbFsetminus 0,cdot)$ 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 $mathbbF^*$ 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 $ein G$ so that for any $gin G$ 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 $qin F$ that $q+0=q$.

A4) For every $q in F$ there is an element $-qin F$ 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 $q in F$, $1cdot q = q$.

[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 $acdot a = a$ and there is absolutely nothing more to say about it). We'll take it for granted that $1 ne 0$]

M4: For every element $qin F$ so that $qne 0$ 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 $Fsetminus 0$ are the same axioms as the axioms of a group with only different notation. So we can state that $F^*$, the set $Fsetminus 0$ associated with the binary operation $cdot$, is a group (with the additional condition $cdot$ is commutative).

The final axiom:

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

......

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 $Fbackslash0$ 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 $(ab)^-1=b^-1a^-1$- 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 $Fbackslash0$. Finally, since $F$ is not the trivial field, $F$ has at least one non-zero element, so $Fbackslash0$ is non-empty. Suppose it has some element, $x$, in it. By the arguments already outlined, it must be the case that $x^-1in F$ and, then, that $x*x^-1=1in F$, so $Fbackslash0$ has a neutral element (again, as the multiplication on $Fbackslash0$ is the same on that on $F$, $1$ is the neutral element of $Fbackslash0$).

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

So $Fbackslash0$ 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 $Fbackslash0$) 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 $y^k-1=0implies y^k=1$ (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 $knot | n$, 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 $mathbb Z_<d>$ 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), $<a>cong mathbb Z_<d>$ (forgive the loose notation). This means that the number of elements in $<a>$ with order equal to exactly $d$ (and note that every element in $G$ with order exactly $d$ must be in $<a>$) is equal to the number of elements in $mathbb Z_<d>$ with order exactly $d$.

(in case you're wondering, $mathbb Z_<d>$ looks like $0,fracnd,frac2nd,...,frac(d-1)nd$)

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 $sum_dG(d)$. Similarly, in $mathbb Z_n$, the total number of elements in all equivalence classes must be $sum_dmathbb Z(d)$. Using the previous paragraph, then $$sum_dG(d)leq sum_dmathbb Z(d)$$

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 $Big|GBig|=Big|mathbb Z_nBig|=n$.

So if for a single $d$, we have $G(d)<mathbb Z(d)$, then we'll have $$sum_dG(d)< sum_dmathbb Z(d)\implies n<n$$

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 $<b>subset G$, but also $operatornameord(b)=n=Big|GBig|$, we must have $<b>=G$ 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.