Q: When is the number of elements in a field equal to the order of its multiplicative group? We know that the multiplicative group $(\mathbb{Q}^*,\cdot)$ is infinite, but is not cyclic. Is it because of the existence of zero divisors? A: The multiplicative group $ (\mathbb{Q}^*,\cdot) $ is indeed infinite and not cyclic because of the existence of zero divisors. The field $ \mathbb{Q} $ is not algebraically closed. Therefore $ \frac{1}{2} $ doesn't have a square root in $ \mathbb{Q} $ but $ \mathbb{Q}(\sqrt 2) $ is a proper subfield of $ \mathbb{Q} $. Since the multiplicative group of a field is a group of automorphisms of the field, we have that if $ |G| = n > 1 $ then $ G \simeq (\mathbb{Z}/n\mathbb{Z})^* $, and that if $ G \simeq (\mathbb{Z}/n\mathbb{Z})^* $ then the group is cyclic. A: This is a typical situation where Galois theory shines. It is not really related to the notion of a field extension. If you take the extension $K = \mathbb Q(i)$, then $K$ is a field and $K^x = \mathbb Q$ for all $x$. In particular, the additive group is infinite cyclic, but the multiplicative group is infinite and not cyclic. The point is that if you had a cyclic group $G$ of order $n$, then any non-trivial Galois extension of $K$ would have order divisible by $n$. The fundamental theorem of Galois theory says that the Galois groups of $K$ are the quotients of the additive group of $K$ modulo its subgroups which are normal in the additive group. Since the additive group is infinite, you can see that there will be plenty of subgroups which are not normal. But all fields $K$ have a unique maximal (by inclusion) proper normal subgroup $K^x$, and in the case of the previous example, the Galois group is $K^*/K^x \cong \mathbb Z$. In general the Galois group is isomorphic to $H^1(G,\mathrm{Aut}(K))$, where $G$ is the Galois group of $K$ over $\mathbb Q$ and $\mathrm{Aut}(K)$ is the group of all automorphisms of $K$ as a field. A: In short, this is because the characteristic of the field being not zero divides all the elements of the group and you can check with some hand work that $\mathbf Z/m\mathbf Z$ is not characteristic zero. As I know, if the characteristic of a field is nonzero, then the group $G=\mathbf Q^{\times}$ cannot be cyclic (and the characteristic is divisible by all $n$ as the group is cyclic). This is related to another common misconception about the characteristic of a field being the dimension of a vector space over $\mathbf Q$. This is a consequence of the above fact and of the isomorphism of the multiplicative group and the group of units (i.e. a quotient of the additive group) of a field. This is all proved in great details in Dummit & Foote's Abstract Algebra. $\mathbf Z/m\mathbf Z$ is not characteristic zero as it has a single characteristic polynomial $X^m-1$ (see this question). The fact that $\mathbf Z/m\mathbf Z$ can't be characteristic zero is already mentioned here: Characterization of Characteristic Zero Field I could say that the condition that the characteristic of the field should not divide the order of a finite group is sufficient but not necessary for the group to be cyclic. In general, let $G$ be a finite group, and let $P(G)$ be its Sylow subgroups. Then $P(G)$ is characteristic zero if and only if the set of order of $G$ divides the set of order of $P(G)$: The fact that all Sylow subgroups are characteristic is proved by Dummit & Foote's Abstract Algebra (see section 9.7, page 858). The fact that the order of the group divides the order of the Sylow subgroups is due to this question, and this theorem proves it. The Sylow subgroup theorem (finite groups) says that $P(G)$ is characteristic zero iff the order of each Sylow subgroup divides the order of $G$ and, in this case, $G$ is called an abelian group. An easy example of a non-abelian group in characteristic zero is $\mathbf Z/4\mathbf Z$ since each subgroup of order $4$ is normal. For a non-abelian finite group, its Sylow subgroups are always normal subgroups, but they are not the only subgroup of a group that are normal.