After classifying some small groups, we now focus on a recurring method: prove that enough Sylow subgroups are unique, then assemble the whole group from them. In this lesson, the focus keyword is cyclicity from Sylow subgroups. The key point is that normal subgroups of relatively prime orders commute elementwise when their intersection is trivial. This turns Sylow information into a direct product decomposition and often proves that the group is cyclic. :::definition[Cyclicity from Sylow Subgroups] Let $G$ be a finite group. We say that $G$ has cyclicity from Sylow subgroups when its Sylow subgroups are normal, cyclic, have pairwise relatively prime orders, and multiply to the whole group. ::: :::lemma Let $H$ and $K$ be normal subgroups of a group $G$. If $H\cap K=\{e\}$, then every element of $H$ commutes with every element of $K$. ::: :::proof Given that $H\trianglelefteq G$, $K\trianglelefteq G$, and $H\cap K=\{e\}$. To prove that every element of $H$ commutes with every element of $K$. Let $h\in H$ and $k\in K$. Consider the commutator $$\begin{equation*} hkh^{-1}k^{-1}. \end{equation*}$$ Since $K$ is normal, $hkh^{-1}\in K$, so $hkh^{-1}k^{-1}\in K$. Since $H$ is normal, $kh^{-1}k^{-1}\in H$, and therefore $hkh^{-1}k^{-1}\in H$. Thus $$\begin{equation*} hkh^{-1}k^{-1}\in H\cap K=\{e\}. \end{equation*}$$ Therefore $$\begin{equation*} hkh^{-1}k^{-1}=e. \end{equation*}$$ This gives $$\begin{equation*} hk=kh. \end{equation*}$$ Hence every element of $H$ commutes with every element of $K$. $\square$ ::: :::theorem Let $G$ be a finite group whose Sylow subgroups are all normal and cyclic, with pairwise relatively prime orders. If their product has order $|G|$, then $G$ is cyclic. ::: :::proof Given that the Sylow subgroups of $G$ are normal and cyclic, have pairwise relatively prime orders, and their product has order $|G|$. To prove that $G$ is cyclic. Let the Sylow subgroups be $P_1,P_2,\ldots,P_k$. Since their orders are pairwise relatively prime, we have $$\begin{equation*} P_i\cap P_j=\{e\} \end{equation*}$$ whenever $i\neq j$. By the preceding lemma, elements from distinct $P_i$ commute elementwise. Therefore the product is an internal direct product: $$\begin{equation*} G=P_1P_2\cdots P_k\cong P_1\times P_2\times\cdots\times P_k. \end{equation*}$$ Each $P_i$ is cyclic, and their orders are pairwise relatively prime. A direct product of cyclic groups of pairwise relatively prime orders is cyclic. Hence $G$ is cyclic. $\square$ ::: This preview shows how normal Sylow subgroups can assemble into a cyclic direct product. Enter a group order or choose a preset from the solved problems. The display lists the Sylow factor orders and checks that they are pairwise coprime. When the corresponding Sylow subgroups are normal and cyclic, these factors commute and their product is cyclic. :::scientific-preview[Normal Sylow Product Builder] ::: The preview identifies the Sylow factor orders but does not prove normality by itself. The written examples use Sylow counting first, then the direct product argument turns the normal cyclic factors into a cyclic subgroup or cyclic group. This calculator checks the direct-product part of the argument. Enter factor orders such as `11,7`, `5,7,19`, or `3,41`. The calculator tests whether the orders are pairwise coprime and whether their product has the expected order. When the factors are cyclic and normal in the surrounding group, the direct product is cyclic. :::calculator[Cyclic Product from Sylow Factors Calculator] ::: The calculator isolates the final algebraic step. Sylow theory supplies normality and subgroup existence; coprime cyclic direct products supply cyclicity. :::solved-problem Let $G$ be a group of order $231=3\cdot 7\cdot 11$. Show that $G$ has a cyclic subgroup of order $77$. ::: :::solution Let $H$ be a Sylow $11$-subgroup of $G$. Then $$\begin{align*} n_{11} &\equiv 1 \pmod {11}, \\ n_{11} &\mid 21. \end{align*}$$ The divisors of $21$ are $1,3,7,21$, and only $1$ is congruent to $1$ modulo $11$. Hence $n_{11}=1$, so $H\trianglelefteq G$. Let $K$ be a Sylow $7$-subgroup. Then $$\begin{align*} n_7 &\equiv 1 \pmod 7, \\ n_7 &\mid 33. \end{align*}$$ The divisors of $33$ are $1,3,11,33$, and only $1$ is congruent to $1$ modulo $7$. Hence $n_7=1$, so $K\trianglelefteq G$. Since $|H|=11$ and $|K|=7$, we have $H\cap K=\{e\}$. Normality gives elementwise commutativity. Thus $$\begin{equation*} HK\cong H\times K\cong C_{11}\times C_7\cong C_{77}. \end{equation*}$$ Hence $G$ has a cyclic subgroup of order $77$. $\square$ ::: :::solved-problem Let $G$ be a group of order $5\cdot 7\cdot 19$. Prove that $G$ is cyclic. ::: :::solution Let $|G|=5\cdot 7\cdot 19$. Let $n_{19}$ be the number of Sylow $19$-subgroups. Then $$\begin{align*} n_{19} &\equiv 1 \pmod {19}, \\ n_{19} &\mid 35. \end{align*}$$ Thus $n_{19}=1$. Let $n_7$ be the number of Sylow $7$-subgroups. Then $$\begin{align*} n_7 &\equiv 1 \pmod 7, \\ n_7 &\mid 95. \end{align*}$$ The divisors of $95$ are $1,5,19,95$, and only $1$ is congruent to $1$ modulo $7$. Hence $n_7=1$. Let $n_5$ be the number of Sylow $5$-subgroups. Then $$\begin{align*} n_5 &\equiv 1 \pmod 5, \\ n_5 &\mid 133. \end{align*}$$ The divisors of $133$ are $1,7,19,133$, and only $1$ is congruent to $1$ modulo $5$. Hence $n_5=1$. Thus all Sylow subgroups are normal. Their orders are pairwise relatively prime and each is cyclic because it has prime order. Therefore $$\begin{equation*} G\cong C_5\times C_7\times C_{19}\cong C_{5\cdot 7\cdot 19}. \end{equation*}$$ Hence $G$ is cyclic. $\square$ ::: :::solved-problem Let $G$ be a group of order $123$. Show that for every positive divisor $d$ of $123$, there is a unique subgroup of $G$ of order $d$. ::: :::solution Since $$\begin{equation*} 123=3\cdot 41 \end{equation*}$$ and $$\begin{equation*} 3\nmid 40, \end{equation*}$$ every group of order $123$ is cyclic by the cyclic criterion for groups of order $pq$. Hence $$\begin{equation*} G\cong C_{123}. \end{equation*}$$ A cyclic group has a unique subgroup of order $d$ for every positive divisor $d$ of its order. Therefore $G$ has a unique subgroup of order $d$ for every positive divisor $d$ of $123$. $\square$ ::: :::exercise 1. Show that every group of order $133$ is cyclic. 2. Let $G$ be a group of order $70$. Prove that $G$ has a cyclic subgroup of order $35$. 3. Explain why normal Sylow subgroups of relatively prime orders commute elementwise. ::: :::answer 1. Since $133=7\cdot 19$ and $7\nmid 18$, every group of order $133$ is cyclic. 2. The Sylow $7$- and Sylow $5$-subgroups are unique and normal. Their product is isomorphic to $C_7\times C_5\cong C_{35}$. 3. If $H$ and $K$ are normal with $H\cap K=\{e\}$, then each commutator $hkh^{-1}k^{-1}$ lies in both $H$ and $K$, so it equals $e$. ::: :::faq[Frequently Asked Questions] Q: Why do normal Sylow subgroups often make a group cyclic? A: Normal Sylow subgroups of relatively prime orders commute and form a direct product; if each factor is cyclic and the orders are coprime, the product is cyclic. Q: Is a product of cyclic groups always cyclic? A: No. The orders must be pairwise relatively prime for the direct product to be cyclic. Q: Why are Sylow subgroups of prime order cyclic? A: Every group of prime order is cyclic. :::