## Cyclic group

Groups, Rings, Domains, Modules, etc, Galois theory
Grigorios Kostakos
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### Cyclic group

Let $G$ a group of order $n$. Prove that if ${\rm{gcd}}(\varphi(n), n)=1$, where $\varphi$ is the Euler totient function, then $G$ must be a cyclic group.
Grigorios Kostakos

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Tolaso J Kos
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### Re: Cyclic group

Hello Grigorios. Shall we push it a bit further?

Prove that an integer $m$ satisfies $\gcd( \psi(m), m)=1$ if and only if every group of order $m$ is nilpotent, where $\psi$ is a multiplicative function defined as:
1. $\psi(1)=1$
2. $\psi(p^v)=(p^v-1)(p^{v-1}-1)\cdots(p-1)$ and $p$ prime as well as $v \geq 1$.
Remarks:
1. The last result is a much stronger one that the result quoted in the first post.
2. The theorem I quote carries with it a name. I'll quote in a later post.
3. In the first post by Grigorios , the converse also holds.
Imagination is much more important than knowledge.
Riemann
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### Re: Cyclic group

Grigorios Kostakos wrote:Let $G$ a group of order $n$. Prove that if ${\rm{gcd}}(\varphi(n), n)=1$, where $\varphi$ is the Euler totient function, then $G$ must be a cyclic group.
$\newcommand{ord}{{\rm ord}}$

In fact both directions hold. That is:

Theorem: Let $n$ be a positive integer. Then every finite group of order $n$ ic cyclic if and only if $\gcd(\varphi(n), n)=1$, where $\varphi$ denotes Euler's totient function.

($\mathbf{\Leftarrow}$) Let $n$ be a positive integer satisfying $\gcd(\varphi(n), n)=1$. Using the canonical form for $\varphi(n)$ , we immediately see that $n$ is square free. We shall prove the result using induction on $n$. Trivially the results holds if $n$ is prime. (that is a known result) [1]. Suppose now that for all positive integers $n' <n$ with $\gcd(\varphi(n'), n')=1$ , all groups of order $n'$ are cyclic. Let $\mathcal{G}$ be a group of order $n$. From the induction hypothesis and Lagrange's theorem , it follows that all proper subgroups of $\mathcal{G}$ are cyclic.

Lemma 1: Any two different elements of a cyclic subgroup of $\mathcal{G}$ are not conjugate.

The proof is left to the reader.

Lemma 2: For each $x \in \mathcal{G}$ with $x \neq 1$, $\mathcal{C}_\mathcal{G} (x)$ is the unique maximal proper subgroup of $\mathcal{G}$ containing $x$.

Lemma 3: Let $\langle \zeta \rangle$ and $\langle \eta \rangle$ be two maximal proper subgroups of $\mathcal{G}$ and let ${\rm d}_1=\ord_\mathcal{G}(\zeta)$ and ${\rm d}_2=\ord_\mathcal{G}(\eta)$. Suppose that $\gcd({\rm d}_1, {\rm d}_2)>1$. Then $\langle \zeta \rangle, \langle \eta \rangle$ are conjugate. In particular ${\rm d}_1={\rm d}_2$.

The proof is left to the reader.

Since the center of $\mathcal{G}$ is trivial , it follows from the class equation that:

$$n=1+\sum_{i=1}^{r}\frac{n}{\left | \mathcal{C}_\mathcal{G} (x_i) \right |'}$$

where $x_i, \; i=1, \dots, r$ are representatives of the different non trivial conjugacy classes. Each of the $\mathcal{C}_\mathcal{G}(x_i)$ are maximal proper subgroups, the order of each two of them being either equal or coprime (Lemma 3). Furthermore, it is clear that for every prime divisor $p$ of $n$ , there exists an $x_i$ such that $p$ divides $\left | \mathcal{C}_\mathcal{G}(x_i) \right |$. Hence $n$ can be written as $n=\prod \limits_{i=1}^{k}n_i$ where each of $n_i$ is the order of some $\mathcal{C}_\mathcal{G}(x_i)$. Notice that $k>1$ and $n_j>1$ for all $j=1, 2, \dots, k$. Every summand in $(1)$ is of the form $\frac{n}{n_j}$ for some $j$. Furthermore , for each $x_i$ , each of the elements $\mathcal{C}_\mathcal{G}(x_i)$ belong to different conjugacy classes by Lemma 1. Moreover, for each $x \in \mathcal{C}_\mathcal{G}(x_i)$ with $x \neq 1$ we have that $\mathcal{C}_\mathcal{G}(x)= \mathcal{C}_\mathcal{G}(x_i)$ since $\mathcal{C}_\mathcal{G}(x_i)$ is a maximal proper subgroup of $\mathcal{G}$ containing $x$, which is unique by Lemma 2. It follows that in the sum in $(1)$ , the summand $\frac{n}{n_j}$ appears at least $n_j-1$ times for all $j=1, \dots, k$. Hence:

$$n \geq 1+ \sum_{j=1}^{k}\frac{n}{n_j} \left ( n_j-1 \right )$$

and so:

$$n \left ( 1+ \sum_{j=1}^{k}\frac{1}{n_j} \right ) \geq 1 + kn$$

Thus:

$$1+ \sum_{j=1}^{k}\frac{1}{n_j} >k$$

and since $n_j \geq 2$ for all $j=1, \dots, k$ this leads to:

$$1+\frac{k}{2} >k$$

so $k<2$, leading us to a contradiction. It thus follows that the center $\mathcal{Z}$ of $\mathcal{G}$ is non trivial. If we let ${\rm d}_1=\left|\mathcal{Z}\right|$ and ${\rm d}_2=\left | \mathcal{G}/ \mathcal{Z} \right |$ we easily observe that ${\rm d}_1 {\rm d}_2=n$ and of course $1<{\rm d}_1 , \; {\rm d}_2 <n$ so by the induction both $\mathcal{Z}$ and $\mathcal{G / Z}$ are cyclic. Let $\zeta$ and $g \mathcal{Z}$ be a generator of $\mathcal{Z}$ and of $\mathcal{G /Z}$ respectively. Let ${\rm d}=\ord_\mathcal{G}(g)$. Then:

$$\left ( g \mathcal{Z} \right )^{\rm d}=g^{\rm d} \mathcal{Z}= \mathcal{Z}$$

This means that ${\rm d}_2 \mid d$. Let $x=\zeta^{\rm d}$. Then:

$$\ord_\mathcal{G}(x)=\frac{{\rm d}_1}{\gcd\left ( {\rm d}_1, d \right )}=\frac{n}{{\rm d}}$$

Assume that $\ord_\mathcal{G}(gx) <n$. Then the order of $gx$ divides $\frac{n}{p}$ for some prime divisor $p$ of $n$. Since $x \in \mathcal{Z}$ ,

$$\left ( gx \right )^{n/p}= g^{n/p} x^{n/p}$$

However $\ord_\mathcal{G}(x) \cdot \ord_\mathcal{G}(g)=n$ and $n$ is square free , so $\frac{n}{p}$ is divisble by exaclty one of the numbers $\ord_\mathcal{G}(x) , \; \ord_\mathcal{G}(g)$. But then either:

$$\left ( gx \right )^{n/p}=g^{n/p} \neq 1 \quad \text{or} \quad \left ( gx \right )^{n/p}=x^{n/p}\neq 1$$

which is a contradiction. Thus $\mathcal{G}$ is cyclic because $\ord_\mathcal{G}(gx)=n$.

($\mathbf{\Rightarrow}$) Let $n$ be a positive integer $n$ satisfying $\gcd(\varphi(n), n)>1$ . We have to show that there exists a finite group $\mathcal{G}$ of order $n$ that is not cyclic. If $n$ is not square free then the result is obvious, because if $p$ is a multiple divisor of $n$ , then the group $\mathbb{Z}_p \times \mathbb{Z}_{n /p}$ is not cyclic.

Suppose now that $n$ is square free and $\gcd(\varphi(n), n)>1$. Then there exist prime divisors $p, \; q$ of $n$ such that $q-1$ is divisible by $p$. Let $g$ be a primite root modulo $q$. We consider the set:

$$\mathcal{H}=\left \{ \sigma:\mathbb{Z}_q \rightarrow \mathbb{Z}_q, \; x \mapsto g^{(q-1)\kappa/p }x +\lambda, \; \kappa, \lambda \in \mathbb{Z} \right \}$$

Clearly , $\left | \mathcal{H} \right |=pq$ since:

$$g^{\left ( q-1 \right )\kappa_1/p}x +\lambda_1 \equiv g^{\left ( q-1 \right )\kappa_2/p}x +\lambda_2 \pmod q$$

for all $x \in \mathbb{Z}$ if and only if $\kappa_1 = \kappa_2 \pmod p$ and $\lambda_1 = \lambda_2 \pmod q$. Furthemore $\mathcal{H}$ is a subgroup of the symmetric group $\mathcal{S}\left ( \mathbb{Z}_q \right )$. Let $\sigma_1$ send $x$ to $x+1$ and $\sigma_2 \in \mathcal{H}$ send $x$ to $g^{\left ( q-1 \right )/p} x$. We can now see that $\mathcal{H}$ is not abelian , because:

$$\left (\sigma_1 \circ \sigma_2 \right ) (x)=g^{\left ( q-1 \right )/p}x +1 \quad, \quad \left ( \sigma_2 \circ \sigma_1 \right ) (x)= g^{\left ( q-1 \right )/p}x + g^{\left ( q-1 \right )/p}$$

Now we consider the group $\mathcal{G}= \mathcal{H} \times \mathbb{Z}_{n/(pq)}$. Since the order of $\mathcal{H}$ is $pq$ , the order of $\mathcal{G}$ is $n$ and $\mathcal{H}$ is not abelian we conclude that $\mathcal{G}$ is also not abelian. In particular $\mathcal{G}$ is not cyclic. With this the exercise comes to an end.

[1] See a proof here.

[2] I have taken this particular proof (for this known theorem) from here.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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