Let $\mathcal{G}$ be a group of order $n$. Prove that if $\gcd(\varphi(n), n )=1$ then $\mathcal{G}$ is cyclic, where $\varphi$ denotes Euler's totient function.
Solution
$\newcommand{ord}{{\rm ord}}$
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). 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.
\begin{equation} n=1+\sum_{i=1}^{r}\frac{n}{\left | \mathcal{C}_\mathcal{G} (x_i) \right |'} \end{equation}
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 ${\rm d}_2 \mid {\rm d}$. 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 $\ord_\mathcal{G}(gx)=n$.
We have taken this particular proof from here
Solution
$\newcommand{ord}{{\rm ord}}$
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). 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.
Proof: It 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 $x$.
Proof: It is left to the reader.
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 ${\rm d}_1={\rm d}_2$.Since the center of $\mathcal{G}$ is trivial , it follows from the class equation that:
Proof: It is left to the reader.
\begin{equation} n=1+\sum_{i=1}^{r}\frac{n}{\left | \mathcal{C}_\mathcal{G} (x_i) \right |'} \end{equation}
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 ${\rm d}_2 \mid {\rm d}$. 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 $\ord_\mathcal{G}(gx)=n$.
We have taken this particular proof from here
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