Establishing additive and multiplicative inverses for a finite field

Question

I am struggling with the following problem:

Let $F$ be a finite field, and let $G$ be a subset of $F$ with the following properties: $0$ and $1$ are in $G$; whenever $a$ and $b$ are in $G$, $a + b$ and $ab$ are in $G$. Prove that $G$ is a field.

So I know that $G$ will inherit the associative, distributive and commutative properties from $F$, meaning that all that remains if to establish that there exists additive and multiplicative identities for all elements in $G$.

Could someone kindly point me in the right direction with this one?

Answer 1

Use the fact that:

If $G$ is a finite group and $H$ be a subgroup of $G$ such that $a,b\in H\implies ab\in H$ then $H$ is a subgroup of $G$ .(Why?)

Only thing that remains to show is that $a\in H\implies a^{-1}\in H$ this is true since the group is finite for any $a\exists n\in \mathbb N$ such that $a^n=1\implies a.a^{n-1}=a^{n-1}.a=1$ .thus $a$ has an inverse

Can you complete now ?Remember a field is a group with both the binary operations

Answer 2

Your reasoning is great so far. Here is a hint for both additive and multiplicative inverses:

If $A$ is a finite group with operation $\star$, then for any given $a\in A$, the inverse of $a$ is a power of $a$; that is, for some $n$, we have $$a^{-1}=a^{n}=\underbrace{a\star a\star\cdots\star a}_{n\text{ times}}$$ That's because any element of a finite group has finite order (e.g., by Lagrange's theorem), so there is some $k\geq 1$ for which $a^k=1$; then we have $$a^{k-1}\cdot a=a^k=1$$ which implies $a^{k-1}$ is the inverse of $a$.

The particular groups to use this observation on would be $(F,+)$ and $(F\setminus\{0\},\,\times\,)$.

Answer 3

A finite (and non-empty) subset $S$ of any group $(G,\ast)$ closed under the group operation $(\ast)$ is a subgroup ($G$ need not be finite). The proof is similar to Zev Chonoles':

Let $a \in S$. By closure, all positive powers of $a$:

$a, a\ast a = a^2, a\ast a\ast a = a^3$, etc. are all in $S$.

Since $S$ is finite, these all cannot be different. Hence there exist two (positive) integers $m < n$, such that $a^m = a^n$. It follows that for $k = n - m$, that:

$a^k = a^{n-m} = a^na^{-m} = a^n(a^m)^{-1} = a^n(a^n)^{-1} = e_G$ (Note here, we are working "in $G$" to prove something about $S$, because we do not yet know if $a^{-m}$ lies in $S$, but it certainly lies in $G$).

Since there is some $k \in \Bbb Z$ with $k > 0$ such that $a^k = e_G$, there must be a least such positive integer (the positive integers are well-ordered). Let's call it $t$.

We have $\{a,a^2,a^3,\dots,a^{t-1}\} \subseteq S$, and these are all distinct, by the minimality of $t$. Clearly, $a^{t-1} = a^ta^{-1} = e_Ga^{-1} = a^{-1}$ (in $G$ - note this is the unique inverse of $a$ in $G$), and thus $a^{-1} \in S$, for any $a \in S$.

Hence $S$ contains all inverses, and consequently the identity $e_G$ as well, and is thus a subgroup of $G$.

If $G$ itself is finite, of course, any subset will be, as well.

Note that $(F,+)$ and $(F-\{0\},\cdot)$ are groups, by definition of a field, and $F$ is finite.

Answer 4

Apply the more general fact below to $G \subseteq (F,+)$ and to $G\setminus0 \subseteq (F\setminus0,\cdot)$.

Let $M$ be a monoid with cancellation. If $S$ is a finite submonoid of $M$, then $S$ is a subgroup of $M$.

We need to prove that every element $a\in S$ has a two-sided inverse in $S$.

Indeed, take $a\in S$ and consider $f: S \to S$ given by $f(x)=ax$. Then:

• $f$ is makes sense because $S$ is closed under the operation of $M$.

• $f$ is injective because $M$ has cancellation.

Therefore, $f$ is surjective because $S$ is finite. In particular, $e \in S$ is in the image of $f$, that is, there is $b \in S$ such that $ab=e$.

By considering $x \mapsto xa$, we conclude in the same way that there is $c \in S$ such that $ca=e$.

Finally, $c=ce=c(ab)=(ca)b=eb=b$, which means that $ab=e=ba$, and $a$ is invertible in $S$.