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In Visual Group Theory by Nathan Carter (Exercise 10.30), the following question is posed:

"Why can the group $C_4$ under addition not be made into a finite field by overlaying a multiplicative structure on $\{1,2,3\}$? Why do $C_6$ and $C_{15}$ have the same problem?"

I understand that a finite field of order $n$ exists if and only if $n = p^k$, where $p$ is a prime and $k \geq 1$. But I’d like to understand more intuitively and structurally why the groups $C_4$, $C_6$, and $C_{15}$, though perfectly fine as additive cyclic groups, cannot be equipped with a multiplication operation to form a field.

azimut
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F. A. Mala
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    I don’t know what you mean. You say you understand why the additive group must have order a prime power. But you don’t understand “intuitively” why these specific groups of non prime power order can’t be the additive group of a field? – Malady Jun 21 '25 at 18:04
  • @Malady This is not necessarily a contradiction. When someone says, "I understand that...", it often means something like "I have it on good authority that...". – user43208 Jun 21 '25 at 18:23
  • @user43208 ahh. I took it to mean: “I have seen and understood a proof of…” – Malady Jun 21 '25 at 18:28

2 Answers2

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A nontrivial finite cyclic subgroup of the additive group of a field $\mathbf F$ must be of prime order: Suppose $0\ne a\in \mathbf F$ and the smallest positive integer $n$ such that adding $n$ copies of $a$ produces $0$ is composite, say $n=rs$ with integers $r,s$ greater than $1$. Let $b$ be the sum of $r$ copies of $a$ and $c$ the sum of $s$ copies of $a$. By minimality of $n$, $b\ne0$ and $c\ne0$. By associativity, we find that $bc$ is the sum of $n$ copies of $a^2$, which is $a$ times the sum of $n$ copies of $a$. By assumption, the latter is $0$, hence $b$ and $c$ are nonzero elements of $\mathbf F$ whose product is $0$. This is not possible in a field!

Since $C_4$, $C_6$, $C_{15}$ are cylic of composite order, they cannot be the additive group of a finite field (they cannot even be a subgroup of the additive group).

Hagen von Eitzen
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    It might be helpful to point out that “$bc$ is the sum of $n$ copies of $a^2$” is where we use distributivity. Since distributivity is really the only obstacle to defining the multiplicative structure given some putative additive structure. – Malady Jun 21 '25 at 18:06
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Let $1$ be a generator of the cyclic group $(C_4,+)$. The neutral element is denoted by $0$. The elements of $C_4$ are $$1,\quad 2 := 1+1,\quad 3 := 1+1+1 \quad\text{and}\quad 0 = 1+1+1+1.$$ Whatever the "overlaying multiplicative structure" of $C_4$ is, the following is forced by the distributive law: $$ 2 \cdot 2 = (1 + 1) \cdot (1 + 1) \overset{\text{Dist.Law.}}{=} 1\cdot (1 + 1) + 1\cdot (1 + 1)\overset{\text{Dist.Law.}}{=} (1\cdot 1+1\cdot 1) + (1\cdot 1+1\cdot 1) = 1\cdot 1+1\cdot 1+1\cdot 1+1\cdot 1= 0. $$ In the last step, we used that $a + a + a + a = 0$ holds for any element $a\in C_4$ and thus, in particular, for $1\cdot 1$. (In this moment, we don't know the result of $1\cdot 1$.)

Hence $2$ is a zero divisor and $C_4$ cannot be made into a field.

For $C_6$ and $C_{15}$, you can conclude similarly that $2\cdot 3 = 0$ and $3\cdot 5 = 0$, respectively.

azimut
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