Finite field where each subgroup of the multiplicative group of nonzero elements of the field, together with $0$, forms a field.

Question

My whole question looks like-

If $F$ is a finite field, then $H\cup\{0\}$ is a subfield of $F$ for each subgroup $H$ of the multiplicative group $F^\ast$ if and only if $|F^\ast|$ is either $1$ or a prime of the form $2^n-1$, where $n$ is a positive integer.

I have just proved the "if" part. Because in that case the subgroups will be $\{1\}$ and $F^\ast$ and hence they form subfield together with $0$. Done.
But can't prove the "only if" part.
Can anybody solve this part of the problem? Thanks for assistance in advance.

Answer 1

Title: "Rudolf Lidl and Harald Niederreiter - Finite Fields"

Ex2.11: Prove that if $F$ is a finite field, then $H\cup\{0\}$ is a subfield of $F$ for every subgroup $H$ of the multiplicative group $F^*$ if and only if the order of $F^*$ is either $1$ or a prime number of the form $2^p-1$ with a prime $p$.

Proof:

Step 1 - $char(F)=2$:

In this case, let's consider $H=\{1\}$, which is a subgroup of any group. Thus, $K := H\cup\{0\}$ must be a subfield of $F$. However, we also know that $|K|=2$, which implies that $|F|=2^n$ for some positive integer $n$.

Step 2 - $2^n-1$ is a prime number:

Now, assume that $2^n-1$ is not prime, and it can be expressed as $2^n-1 = m \cdot t$, where $m$ and $t$ are positive integers greater than $1$. Since every $F^*$ is a cyclic group, let's choose an element $a \in F^*$ as a generator, such that $F^*=\langle a \rangle$. The subgroups $H_1=\langle a^t \rangle$ and $H_2=\langle a^m \rangle$ are now subgroups of $F^*$, and their orders are $|H_1|=m$ and $|H_2|=t$. Consequently, the sets $H_1\cup\{0\}$ and $H_2\cup\{0\}$ are subfields of $F^*$.

This leads us to the conclusion that $|H_1|=2^r$ and $|H_2|=2^k$ for some positive integers $k$ and $r$. As a result, we have:

$$2^n-1 = m \cdot t = (2^r-1) \cdot (2^k-1)$$

Then, by simplifying:

$$2^n-1 = 2^{r+k}-2^r-2^k-1$$

Further simplifying:

$$2^n-2^{r+k}-2^r-2^k = -2$$

If both $r$ and $k$ are greater than $1$, the left side of the equation is divisible by $4$. So, at least one of them must be equal to $1$. For example, if we assume $r=1$, then $m=1$, and $2^n-1=m\cdot n=1\cdot n$. Thus, in the case when $r=k=1$, we have $|F^*|=2^n-1=1$.

Step 3 - $n$ is not prime:

If $n$ is not prime, we can express it as $n = m \cdot t$, where both $m$ and $t$ are greater than $1$. In this case, we have:

$$2^n-1 = 2^{m \cdot t} - 1$$

And since $2^m-1$ is a factor of $2^{m \cdot t} - 1$, we can conclude that $2^n-1$ is divisible by $2^m-1$, contradicting our assumption that it is a prime number.

Answer 2

Assume that the subgroup property holds, but $|F^*|=ab$ is composite with $a,b\ge 2$, and $p:=\operatorname{char}(F)> 2$. Then the cyclic group $|F^*|$ has a subgroup of order $a$, hence we should have a subfield of order $a+1$. For this $a+1$ and $ab+1$ need multiples of $p$ and hence also $b(a+1)-(ab+1)=b-1$ must be a multiple of $p$. By the same reasoning, both $b+1$ and $a-1$ must be multiples of $p$. But then $(a+1)-(a-1)=2$ is a multiple of $p$, contradiction.

Answer 3

If $|H|=k$, then $k+1\mid p^n$ and $k\mid p^n-1$, so we can write:

$p^n=(k+1)m\tag{1}$ $p^n-1=kl\iff p^n=kl+1\tag{2}$

$(2)-(1)$ gives:

$$1=(k+1)m-kl\tag{3}$$

The only solution to $(3)$ is $m=l$, since the difference is between consecutive integers. So in all,

$p^n=(k+1)m\tag{4}$

$p^n=km+1\tag{5}$

$(5)-(4)$ gives $m=1$. Can you continue from here?