Collecting proofs that finite multiplicative subgroups of fields are cyclic

Question

I teach elementary number theory and discrete mathematics to students who come with no abstract algebra. I have found proving the key theorem that finite multiplicative subgroups of fields are cyclic a pedagogical speedbump. For example, Serre's proof in A Course in Arithmetic runs a full page, requires introducing Euler's $\phi$-function, and depends on a counting argument that might seem to beginners too clever or magical for a cornerstone result.

I'd like to have a collection of proofs of this fact, to compare their advantages, to match their viewpoints to my various audiences, to contrast for my students, etc.

To get the ball rolling, here's the shortest argument I can think of (and if it's in the literature somewhere I'd love a reference).

Induction on the order of the subgroup. So suppose multiplicative subgroup $G$ of field $F$ has order $n$. If $n=p^k$ with $p$ prime and $G$ isn't cyclic, all $p^k$ elements of $G$ satisfy $x^{p^{k-1}}-1=0$, impossible.
If $n=ab$, $\gcd(a,b)=1$, then $(\cdot)^a:G\rightarrow G$ has a kernel $A$ of size at most $a$ and a range $B$ of size at most $b$ (since the $y\in B$ satisfy $y^b=1$), so $|A|=a$, $|B|=b$, and a product $xy$ of cyclic generators $x,y$ for $A,B$ respectively generates $G$.

If you know published proofs distinctly different from either of these, please cite a source. No need to spell out the details, but please mention a key feature to help avoid duplicates. If you have your own favorite approach, please share it.

Answer 1

Let $n = |G|$ and let $m$ be the l.c.m. of the orders of the cyclic factors of $G$. Then $x^m = 1$ for all $x \in G$; since we are in a field this equation has at most $m$ roots, which shows that $m \geq n$. It follows that $m = n$ and $G$ is cyclic.

Of course here one uses the classification of finite abelian groups as product of cyclic groups, which you may want to avoid.

Answer 2

I once collected six [edit: now seven [edit: now eight [edit:now nine [edit: now ten]]]] proofs of this theorem, for the field $\mathbf Z/(p)$, and they can be found here. While $\mathbf Z/(p)$ is not a general finite field, since the intent of this MO question is to use proofs in a course for undergraduates without much background I think surely $\mathbf Z/(p)$ is the only finite field that matters for that pedagogical purpose.

Answer 3

Let $G$ be a finite subgroup of $F^{\ast}$ of order $n$. Then all the elements of $G$ satisfy $x^n = 1$ in $F$. Since polynomials of degree $n$ over a field have at most $n$ roots, it follows that the roots of $x^n = 1$ in $F$ are precisely the elements of $G$.

The intuitive content of Serre's argument is as follows: if no element of $G$ has order $n$, then they all have to have order less than $n$, so they satisfy various smaller polynomials $x^d = 1$ for $d | n$, and what the counting argument is trying to show is that there isn't enough "room" in these polynomials for all of these roots. I think this is quite intuitive, and it is completely clear for $n$ a prime power, but you want to avoid it, so:

Over $\mathbb{C}$, the roots of $x^n = 1$ are precisely the $n^{th}$ roots of unity. It is natural to organize these by their order, so let $\Phi_d(x) = \prod_{\zeta \text{ has order exactly } d} (x - \zeta)$. The result that Serre is trying to avoid with his counting argument is that $\Phi_d(x)$ has integer coefficients, so the factorization

$$x^n - 1 = \prod_{d | n} \Phi_d(x)$$

makes sense over an arbitrary field. If you can show this, the rest of the proof is trivial: since $x^n = 1$ splits over $F$, it follows that $\Phi_n(g) = 0$ for some $g \in G$, and such an element must have order $n$ and therefore be a generator.

If your students really have no algebra background I think you should consider stating this without proof. It is easy to give examples and hopefully you can give enough to convince them.

The shortest way I can think of to prove that $\Phi_n(x)$ has integer coefficients is by induction and the identity $\gcd(x^n - 1, x^m - 1) = x^{\gcd(n,m)} - 1$, which again 1) is intuitive over $\mathbb{C}$ but 2) makes sense over an arbitrary field. But this is a bit of a detour and precisely why Serre did something trickier. However, I think the larger lesson that "algebraic things that are intuitive over $\mathbb{C}$ are worth generalizing" is worth learning.

Answer 4

Let $n$ be the number of elements of $F^*$, $p$ be a prime dividing $n$, $q$ be the largest power of $p$ dividing $n$; let $r=q/p$. Look at the map $x \mapsto x^{(n/q)}$, $F^*\to F^*$. The kernel has order at most $n/q$, so the image has order at least $q$, and there are at least $q$ solutions of $x^q=1$. Since there are at most $r$ solutions of $x^r=1$, there is an element of exact order $q$; multiplying these elements together for the various $p$ dividing $n$ gives a generator.

(I've used this no doubt well-known argument successfully in undergrad courses).

Edit: For the final step, let $u$ be the product. Then $u^{(n/p)}=a^{(n/p)}$ where $a$ has exact order $q$. So $u^{(n/p)}$ is not $1$ for all $p$, and $u$ has exact order $n$, and is a generator. Looking again at the question, I realize that this is essentially the same as the proposer's short solution, though I've restricted my attention unnecessarily to finite fields. But it combines Lagrange's theorem with the theorem that $x^m=1$ has at most $m$ solutions in $F^*$ in a very simple way.

Answer 5

I like the following explanation of the fact that if $G=\{g_0,\dots,g_{n-1}\}$ is your group, and $0<m<n$, then there exists $i$ for which $g_i^m\ne 1$:

otherwise the Vandermonde matrix $(g_i^j)$ would have two equal columns, thus zero determinant, but its determinant is $\prod (g_i-g_j)\ne 0$.

(After that we continue in usual way, with primary factors.)

The difference with the usual argument (the polynomial $x^m-1$ has degree $m$, hence at most $m$ roots) is that we replace this general fact by implementing its proof for a specific polynomial which we need. It may have some advantage in other questions around. For example, it allows to prove (by literally the same reasoning) the following generalization:

Let $R$ be a commutative ring with unity $e$. Assume that $G$ is a finite multiplicative subgroup of $R$ and for any $g \in G$, $g \ne e$, the element $e-g$ is not a zero divisor in $R$. Then group $G$ is cyclic.

Answer 6

Lemma: Let $G$ be a finite abelian group, and let $x\in G$ with maximal order. Then for any other element of $y\in G$, $|y|$ divides $|x|$.

Proof. If not, then there is an element $y\in G$ and a prime $p$ such that
$$ |x|= p^a u \qquad |y| = p^b v $$ with $(p,u)=(p,v) = 1$ and $a< b$. In this case, we have $|x^{p^a}| = u$ and $|y^v| = p^b$. Since these orders are relatively prime (and the group is abelian),
$$ |x^{p^a}y^v| = |x^{p^a}|\cdot |y^v| = p^b u > p^au = |x|. $$

---

Now apply this to our subgroup $G\subseteq F^*$. Let $x\in G$ have maximal order $n$, so that $n \leq |G|$. The Lemma implies that there are at least $|G|$ roots to the polynomial $x^n-1$ in the field $F$, and of course there are at most $n$ roots to $x^n - 1$. Thus $$ n \leq |G| \leq \fbox{number of roots of $x^n-1$} \leq n . $$

Answer 7

In some sense this is similar to Andrea Ferretti's post and the comments below it, but with perhaps slightly different phrasing. (It is the proof I wrote up for the nLab some time back.)

As noted, if $G$ is a finite subgroup of $K^\times$, then the exponent $e$ of $G$ (the lcm of orders of its elements) equals the order $m$ since $\prod_{g \in G} (x - g)$ divides $x^e - 1$. So it remains to show that $e = m$ forces $G$ to be cyclic, i.e., that some element has order $e$.

Let $e = p_1^{r_1} p_2^{r_2} \dotsm p_k^{r_k}$. The exponent $r_i$ is the maximum multiplicity of $p_i$ occurring in orders of elements; any element realizing that maximum has order divisible by $p_i^{r_i}$, and then some power $y_i$ of that element has order exactly $p_i^{r_i}$. By the following lemma and induction, $\prod_{i=1}^k y_i$ has order $e$.

Lemma: If $m, n$ are relatively prime and $x$ has order $m$ and $y$ order $n$ in an abelian group, then $xy$ has order $mn$.

Proof: Suppose $(x y)^k = x^k y^k = 1$. For some $a$, $b$ we have $a m - b n = 1$, and so $1 = x^{k a m} y^{k a m} = y^{k a m} = y^k y^{k b n} = y^k$. It follows that $n$ divides $k$. Similarly $m$ divides $k$, so $m n = \operatorname{lcm}(m, n)$ divides $k$, as desired.

Answer 8

I know less than you know in the topic since you are a teacher now.
However, I want to mention two sources you can find the proof which use little prerequisites of algebra.
First of all, the classic Basic Number Theory by André Weil contains a proof in the first section of the first chapter which uses a great method.
As for the second, the Chinese mathematician Hua, Lo-keng (in Chinese: 華羅庚) has published a book entitled Introduction to number theory, which has a proof that uses only elementary techniques, and I hope it is exactly what you need.
By the way, the first approach is the same as that mentioned by @QiaoChu Yuan in some sense, and the second is mostly elementary.

Answer 9

I don't know how helpful this is for anybody, especially students, but for finite subgroups $G$ of $\mathbb C^*$ you can first observe that every element has modulus $1$, so is on the unit circle and has rational argument, and then choose the element $z$ of least non-zero argument. Then, given $y\in G$, rotate clockwise by dividing by powers of $z$ until the argument lies below that of $z$; this shows that $y$ is a power of $z$.

Answer 10

I actually think it will not be so easy to say when two proofs of this result will be "distinctly different": rather I expect most or all will have common features, including using at least a little bit of group theory.

For instance, the proof I wrote up for my elementary(ish) number theory course is Theorem 9 in these notes. The notes themselves are on finite commutative groups, and Theorem 9 is on page 3, in the section on "cyclic groups". Prior to the statement and proof, a little over a page is spent developing the basic properties of cyclic groups, including a statement involving the Euler $\varphi$-function. The proof of the result itself -- which, note, is a criterion for an a priori noncommutative finite group to be cyclic -- occupies $11$ lines. (Added: sorry, false advertising -- add two more lines to get from Theorem 9 to Corollary 10, which is the statement that any finite subgroup of the multiplicative group of a field is cyclic.) I certainly think it is more or less the proof that any research mathematician is expecting to find.

Let me mention though that I had originally included this argument as an application of the Mobius Inversion Formula. After having looked back at what I'd done, I decided that although the argument was reminiscent of an inversion / inclusion-exclusion counting argument, it only made it more complicated to phrase it in that way.

Answer 11

Weil: "Basic Number Theory"

Chapter I

Lemma 1. If K is a commutative field, every finite subgroup of Kˣ is cyclic.

In fact, let Γ be such a group, or, what amounts to the same, a finite subgroup of the group of all roots of 1 in K. For every n ≥ 1, there are at most n roots of xⁿ = 1 in K, hence in Γ; we will show that every finite commutative group with that property is cyclic.

Let α be an element of Γ of maximal order N. Let β be any element of Γ, and call n it's order. If n does not divide N, there is a prime p and a power q = pᵛ of p such that q divides n and not N. Then one verifies at once that the order of αβ^n/q is the l.c.m. of N and q, so that it is > N, which contradicts the definition of N. Therefore n divides N. Now xⁿ = 1 has the n distinct roots α^kN/n in Γ, with 0 ≤ k < n; as β is a root of xⁿ = 1, it must be one of these. This shows that α generates Γ.

Answer 12

Here's my version of a proof, which seems to me to be very elementary. I'll prove the following assertion, which clearly applies to finite subgroups of the multiplicative group of a field:

Let $G$ be a finite abelian group such that, for every prime $p$, there are at most $p$ elements $g\in G$ satisfying $g^p=1$. Then $G$ is cyclic.

I argue by induction on $n=|G|$. So let $G$ be a non-trivial group satisfying the hypothesis. Then there exists an element of $G$ with order $>1$, and hence (by taking an appropriate power) an element of $G$ with order $p$ for some prime $p$. The main observation I will use is that the $p$th power map on $G$ is a homomorphism, and therefore defines a $p$-to-$1$ surjective function $\phi\colon G\to G^p$ (as the hypothesis on $G$ implies that the kernel of $\phi$ must have order $p$.)

The subgroup $G^p$ has order $n/p<n$, and being a subgroup of $G$ also satisfies the hypothesis, so by induction is cyclic on some generator $a\in G^p$. I will show the following: there exists $b\in G\smallsetminus G^p$ such that $b^p=a$. Given this it is straightforward to verify that $b$ has order $n$ (using that $G/G^p$ has prime order so is cyclic), whence $G$ is cyclic.

There are two cases, depending on the behavior of the restriction of the $p$th power map to $G^p$, which gives a surjective homomorphism $\phi|_{G^p}\colon G^p\to G^{p^2}$.

1. $G^{p^2}=G^p$, so $\phi|_{G^p}\colon G^p\to G^p$ is a bijection. Therefore the restriction $\phi|_{G\smallsetminus G^p}$ of the $p$th power map to the complement of $G^p$ is a $(p-1)$-to-$1$ surjective function $G\smallsetminus G^p\to G^p$. In particular $a=b^p$ for some $b\in G\smallsetminus G^p$ as desired.

2. $G^{p^2}\neq G^p$. Therefore $\phi|_{G^p}\colon G^p\to G^{p^2}$ is a surjective $p$-to-$1$ map. Since $\phi\colon G\to G^p$ is surjective and $p$-to-$1$, it follows that the restriction $\phi|_{G\smallsetminus G^p}$ of the $p$th power map to the complement of $G^p$ is a $p$-to-$1$ surjective function $G\smallsetminus G^p\to G^p\smallsetminus G^{p^2}$. The generator $a$ of $G^p$ cannot be in the proper subgroup $G^{p^2}$, so $a=b^p$ for some $b\in G\smallsetminus G^p$ as desired.

Answer 13

Here is a proof that $\mathbb F_q^\times$ is cyclic. Let the finite extension $F/\mathbb Q_p$ have residue field $\mathbb F_q$. The group $\mu_{q-1}=\{x\in F:x^{q-1}=1\}$ is isomorphic to $\mathbb F_q^\times$ by Hensel's lemma.

By choosing an embedding $F\hookrightarrow\mathbb C$, the group $\mu_{q-1}$ can be identified with $\{x\in\mathbb C:x^{q-1}=1\}$. But this is cyclic with generator $e^{2\pi i/(q-1)}$.

Answer 14

This approach goes along elementary ideas. But the proof using the power group homomorphism and the size estimate of the kernel and the image using number of the polynomial roots is much more elegant and simple.

The number $m$ of elements of a subgroup $H$ divides the number $n$ of elements of a group $G$. It simply follows by freely acting by translation with the subgroup on the group (number of the orbits times the number of elements in the subgroup gives the number of elements in the group). The order $k$ of any element is the size of the subgroup generated by that element, therefore $k$ divides $n$. In particular, for any $g$ in $G$, $g^n=1$ (well known!).

Lemma 1. Let $a$ be an element of order $u$ and $b$ an element of order $v$ in $G$ (commutative). If nothing but $1$ divides both $u$ and $v$, then:

(I) the intersection of the two groups generated by $a$ and $b$ contains $1$ and nothing more.

(II) the order of the product $a b$ equals the product $u v.$

Proof: (I) the intersection being a subgroup of both cyclic groups, the number of the elements of the intersection should divide both $u$ and $v$.

(II) $(a b)^{u v}= 1 = (a^u)^v (b^v)^u$, hence the order $l$ of $a b$ is a divisor of $u v$. Also $a^l b^l =1$, hence $a^l=b^{-l}$. From (I) above we get $a^l=1$ and $b^l=1$; hence $u$ and $v$ divide $l$. Therefore $l = u v$.

Lemma 2. If $a$ has order $u$ and $u = w t$ then $c=a^w$ has order $t$ . (obvious).

Using the above results I want to show that:

Prop 1. In a commutative group that is not cyclic there are two distinct cyclic subgroups of the same order (with some elements in both).

Corollary. In a commutative group that is not cyclic, for some integer $w$ there are $w+1$ distinct elements $h$ with the property $h^w=1$.

Proof: Suppose $a$ is chosen in $G$ such it has the highest possible order $m$ and $m < n$ (the number of elements in $G$). Take $b\neq a^i$, i.e. $b$ is not in the group generated by $a$; suppose $b$ has order $k$. We show that $k$ divides $m$. If $k$ has a divisor $p^j$ ($p$ prime) that does not divide $m$ (this means either $p$ does not divide $m$ or the power $j$ is higher than the power of $p$ dividing $m$), using the results above we can build a new element having order $w >= p m$ (first, using lemma 2, raise $a$ to a power that takes out $p$ from its order, second raise $b$ to a power that takes out all primes except $p$ from its order and finally, using lemma 1(II), multiply the two elements), which contradicts how $a$ was chosen.

Therefore $k$ divides $m$ and $a_0=a^{m/k}$ generates a group of order $k$ as $b$ does. Moreover $b$ is not in that subgroup. These are the subgroups looked for.

Prop 2. Suppose $G$ is a finite group (with respect to multiplication) inside a field. Then $G$ is cyclic.

If $G$ is not cyclic, using Corollary above we get an integer $w$ for which $X^w-1$ has $w+1$ roots. Since we are in a field, this is impossible, as for every root $r$ of a polynomial we get a factor $(X-r)$ of that polynomial. And a polynomial of degree $d$ cannot have more than $d$ factors.

Answer 15

The multiplicative group of a finite field $F$ with $n$ elements has at most $d$ solutions of $x^{d} = 1$ for each divisor $d$ of $n.$ It is proved ( or maybe is an exercise) in Herstein's "Topics in Algebra" that a finite group $G$ of order $n$ which has at most $d$ solutions of $x^{d} = 1$ for each divisor $d$ of $|G|,$ then $G$ is cyclic (it is not assumed that $G$ is Abelian). Later edit: Here is a proof of this fact which occurred to me six years later. It is not much more difficult in the non-Abelian case, but depending on the intended audience, it can be simplified just to deal with the Abelian case. I have deliberately avoided using Sylow's Theorem, though it can be much simplified in Sylow's Theorem is allowed to be used. Let $G$ be a finite group of order $n$ which contains at most $d$ solutions of $x^{d} = 1$ for every divisor $d$ of $n$. We wish to prove that $G$ is cyclic.

By induction, we may suppose that every proper subgroup of $G$ is cyclic. Next, we note that every subgroup of $G$ is normal in $G$. Let $H$ be a proper subgroup of $G$. Then $H$ contains $|H|$ solutions of $x^{|H|} = 1$, so $H$ is exactly the set of solutions of $x^{|H|} = 1$ in $G$. But for any $g \in G$, $g^{-1}Hg$ is another subgroup of $G$ of order $|H|$ which also contains $|H|$ solutions of $x^{|H|} = 1$. Hence $g^{-1}Hg = H$ and $H \lhd G$.

Now whenever $A,B$ are subgroups of $G$, we have $AB = BA$ and so $AB$ is a subgroup of $G$ of order $\frac{|A||B|}{|A \cap B|}.$ A proper subgroup $H$ of $G$ is said to be maximal if $H$ is not strictly contained in any proper subgroup of $G$.

Now every maximal subgroup of $G$ has prime index. For let $H$ be a maximal subgroup of $G$ and let $x \in G-H$ be of minimal order. Then $x \neq 1,$ so choose a prime divisor $p$ of the order of $x$. Than $x^{p} \in H$, so that $G = H\langle x \rangle$ and $|G| = \frac{|H||\langle x \rangle|}{|\langle x^{p} \rangle |} = p|H|.$

Now $H$ is cyclic, so has a subgroup of order $d$ for each divisor $d$ of $|H|$. Let $p^{r}$ be the largest power of $p$ dividing $|H|$. Then $p^{r} \geq |\langle x^{p} \rangle|$ and $H$ has a unique subgroup $K$ of order $p^{r}.$ If $p^{r} > |\langle x^{p} \rangle|$ then $x \in K$ since $G$ has a unique subgroup of order $p^{r}.$ This contradicts $x \not \in H.$ Thus $p^{r} = \frac{|\langle x \rangle | }{p}.$

Let $L$ be a subgroup of $H$ of order $\frac{|H|}{p^{r}}$ which we now know is coprime to $p$. Then $|G| = |L| |\langle x \rangle|$ and $L \cap \langle x \rangle = 1$. Since $L \lhd G$ and $\langle x \rangle \lhd G$, we have $G \cong L \times \langle x \rangle$ which is cyclic (generated by $\ell x$, where $\ell$ is a generator of $L).$

Answer 16

First off, this only requires $F$ to be an integral domain, and not because one can form its field of fractions, but because division is never used (other than Euclidean division by monic polynomials in $F[X]$). Like virtually all proofs listed, this one hinges on two statements, which I'll formulate as limited/specific as possible: (1) for any $d>0$ there are at most $d$ distinct elements $x\in F$ satisfying $x^d=1$, and (2) a finite group that has at most one cyclic subgroup of any given order is itself cyclic. I will need some really easy facts that I won't stoop down to prove here: substitution $X:=x$ defines a ring morphism $F[X]\to F$ for any $x\in F$ sending any polynomial to the remainder in its Euclidean division by $X-x$; any element of $G$ generates a cyclic subgroup of$~G$ that has order dividing the order $n$ of $G$ (an instance of Lagrange's theorem); and finally cyclic groups of any positive order$~m$ exist, they are all isomorphic, and they have exactly one subgroup of order$~d$ whenever $d$ divides$~m$. Apart from this I will try to be really elementary; I won't mention Euler's totient function by name (but it's there), and I don't think I even use unique factorisation of integers.

Clearly (1) and (2) combine to show $G$ is cyclic: $G$ must satisfy the hypothesis of (2), since more that one cyclic subgroup of order $d$ will produce (in their union) more than $d$ solutions to $x^d=1$.

To prove (1), first note that in $F[X]$ the leading term of any product of non-zero elements $P,Q$ is the product of their leading terms, so $\deg(PQ)=\deg(P)\deg(Q)$, and if $PQ$ is monic we can modify $P$ and $Q$ by invertible scalars if necessary to make them monic as well. One can then (by induction on the degree) decompose $X^d-1=P_1\ldots P_k$ as a product of monic non-constant irreducible factors $P_i$ (irreducibility being the impossibility to decompose further). By degree consideration $k\leq d$. For $x\in F$ the evaluation $X:=x$ produces a product in $F$ of the remainders of the $P_i$ divided by $X-x$, a remainder that is zero only if $P_i=X-x$. Since $F$ is an integral domain, this must happen for at least one $i$ for every $x\in F$ with $x^d=1$, and this can happen for at most $k\leq d$ different values$~x$.

To prove (2), compare $G$ with the cyclic group $C$ of same order$~n$. By the hypothesis (and mentioned prerequisites) $G$ contains no more cyclic subgroups of any given order than $C$ does. Since the number of generators of a cyclic subgroup depends only on the order of that subgroup, $G$ contains no more elements of any given order than $C$ does. But the fact that $G$ and $C$ both have $n$ elements then implies that we have equalities all along; in particular $G$, like $C$, has a cyclic subgroup of order$~n$, which of course is $G$.

Answer 17

There is a quite elementary proof that uses only the fact that a polynomial of degree $n$ has at most $n$ roots in a field.

So, let $F$ be a field and $G$ be a finite subgroup of the multiplicative group $F^*:=F\setminus\{0\}$ of $F$. To prove that $G$ is cyclic, choose an element $g\in G$ of maximal possible order $n$ in $G$. Let $H:=\{g^k:0\le k<n\}$ be the cyclic subgroup of $G$, generated by the element $g$. If $H=G$, then the group $G$ is cyclic and we are done. So, assume that $H\ne G$. Observe that every element $x\in H$ satisfies the equation $x^n-1=0$, which has at most $n$ roots in the field $F$. Therefore, $H=\{x\in F:x^n=1\}$. Let $p$ be the smallest positive number for which there exists an element $y\in G\setminus H$ such that $y^p\in H$. The minimality of $p$ implies that $p$ is a prime number. Let $k\in\{0,\dots,n-1\}$ be the smallest possible number such that $y^p=g^k$ for some element $y\in G\setminus H$. We claim that $k<p$. Assuming that $k\ge p$, we can observe that the element $z=yg^{-1}\in G\setminus H$ has $z^p=(yg^{-1})^p=g^{k-p}$, which contradicts the minimality of $k$. If $k=0$, then $y^p=g^k=1$ and $y\notin H$ imply that $p$ does not divide $n$. In this case the element $yg\in G$ has order $pn>n$, which contradicts the maximality of $n$. This contradiction shows that $k>0$. It follows from $0<k<p$ that the element $g^k$ has order $d>\frac{n}{p}$. Then the element $y$ has order $pd>n$, which contradicts the maximality of $n$. This is a final contradiction showing that the group $G=H$ is cyclic.

Answer 18

As for me a thorough answer would be to consider the characteristic of the field $F$.

If $\operatorname{char} F>0$, then the finite group $G$ will be contained in a finite field $\mathbb{F}_q$, whose multiplicative group is cyclic, then so is its subgroup.

If $\operatorname{char} F=0$, then $G$ is contained in $\overline{\mathbb{Q}}$ and an element is of form $e^{\frac{q}{p}2\pi i}$ where $\frac{q}{p}\in\mathbb{Q}/\mathbb{Z}$. So it suffices to show that a finite subgroup $H$ of $\mathbb{Q}/\mathbb{Z}$ is cyclic. If $H=\{\frac{q_1}{p_1},\dotsc,\frac{q_n}{p_n}\}$, then it's a subgroup of the cyclic group generated by $\frac{1}{\prod\limits_{i=1}^np_i}$ hence is cyclic.

Answer 19

For the field ${\bf{Z}}/(p)$, we can argue this way. Suppose there is $K\le({\bf{Z}}/(p))^\times$ such that $K\cong C_q\times C_q$, for some prime $q$. Therefore, there are $q^2$ elements $x\in({\bf{Z}}/(p))^\times$ such that $x^q\equiv 1\pmod p$: a contradiction, because this equation has at most $q$ solutions in ${\bf{Z}}/(p)$. So, $({\bf{Z}}/(p))^\times$ does not contain any subgroup isomorphic to $C_q\times C_q$, for any prime $q$. For this characterization of the non-cyclic finite abelian groups, $({\bf{Z}}/(p))^\times$ is cyclic.