Cyclicity of group of order $455$

Question

The following excerpt is from my lecture notes and I have understood almost everything except one moment.

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Prove that any group of order $5\times 7\times 13$ is cyclic.

Using Sylow's theorem we get that $n_7=1$, $n_{13}=1$ and $n_5=1$ or $13\times 7=91$. Suppose $H$ and $K$ are Sylow $7$ and $13$-subgroups of $G$. Then $H\vartriangleleft G$ and $K\vartriangleleft G$ with $o(H)=7$ and $o(K)=13$.

It's now easy to show that $G=HKJ$, where $J$ is a subgroup of order $5$. Since $H=\langle h\rangle$, $K=\langle k\rangle$ and $J=\langle j\rangle$. Then elements of $G$ has form $h^ak^bj^c$. We want to show that $G$ is abelian group because if we can show it then any subgroup will be normal and hence there exists unique Sylow $5$-subgroup and after some machinery work it follow that $G$ is cyclic.

In order to show $G$ is abelian, it is enough to show that elements $h,k,j$ commutes with each other, i.e. $hk=kh$, $kj=jk$ and $hj=jh$.

Since $o(HJ)=7\times 5=35$, $o(KJ)=13\times 5=65$ and $o(HK)=7\times 13=91$ and using theorem about groups of order $pq$ we get that each of these groups are cyclic.

Conclusion: $h,k,j$ all commutes with each other.

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The only thing which I was not able to grasp how from cyclicity follows that for example, $hk=kh$? Sorry if this question sounds dumb. I have tried different ways but no results.

Would be very grateful for explaining me my question!

Answer 1

There are numbers $n$ such that every group of order $n$ is cyclic. These are called cyclic numbers and are characterized by $\gcd(n,\phi(n))=1$, where $\phi$ is Euler's function. See also this question.

Since $\phi(455)=288= 2^5 \cdot 3^2$, we have $\gcd(455,\phi(455))=1$, and so every group of order $455$ is cyclic.

Answer 2

Let $H$ be the $13$ Sylow subgroup of $G$. But we already know $H$ is normal in $G$, so $$N(H)=G$$ where $N(H)$ denotes the normalizer of $H$. But $N(H)/C(H)$ is isomorphic to a subgroup of $\text{Aut}(H)$, so $\vert N(H)/C(H) \vert $ divides $\vert \text{Aut}(H) \sim U(13)\vert=12.$ Also $\vert N(H)/C(H) \vert $ divides $455$, so $$\vert N(H)/C(H) \vert=1 $$ since divisors of $455$ are $1, 5, 7 ,13, 35, 65, 91, 455$.

So $C(H)=G$. Which means every element of $G$ commutes with every element of $H$ and hence $H \subset Z(G)$. Now $13$ divides $\vert Z(G) \vert$ and $\vert Z(G) \vert$ divides $455$. So $$\vert Z(G) \vert \in \{13,35,65,91,455\}$$ and so $$\vert G/Z(G) \vert \in \{35,13,7,5,1\}$$ But the groups of order $35,13,7,5,1$ are cyclic, $G/Z(G)$ is cyclic and so $G$ is Abelian and by Fundamental theorem of finite Abelian groups, $G$ is cyclic!

Answer 3

Let $P$ be the $7$-Sylow subgroup. Since $P$ is normal in $G$, it must be the union of conjugacy classes of $G$; these latter have sizes dividing the order of $G$ (orbit-stabilizer): so, either $1$, or $5$, or higher values amongst $7$, $13$, $5\cdot7$, etc.. Since $\left|P\cap Z(G)\right|\ge 1$, necessarily: $$7=\left|P\cap Z(G)\right|+5k \tag 1$$ for some nonnegative integer $k$. But $\left|P\cap Z(G)\right|=1$ or $7$, so the only possible option is $P$ being central (and $k=0$). Moreover, $G/P$ is cyclic, because $5\nmid (13-1)$. Therefore, by the usual "$G/Z(G)$-argument", $G$ must be Abelian, and hence cyclic (take the product of any three elements of orders $5$, $7$ and $13$, respectively).