Having established the Sylow theorems, the next natural question is how they help us detect the internal shape of a finite group. One of their most powerful uses is to prove that a group is not simple. The guiding idea is direct: if Sylow congruence and divisibility force a Sylow subgroup to be unique, then that Sylow subgroup is normal, giving a nontrivial proper normal subgroup. In this lesson, the focus keyword is simple groups and non-simplicity tests, and the goal is to turn Sylow counting into a practical test for non-simplicity. Students often make the mistake of stopping after finding a subgroup; the subgroup must be normal before it proves non-simplicity. :::definition[Simple Group] Let $G$ be a group such that $G \neq \{e\}$. Then $G$ is called a simple group if the only normal subgroups of $G$ are $\{e\}$ and $G$. ::: A simple group is not a group with a simple multiplication table. It is a group that cannot be broken by quotienting through a nontrivial normal subgroup. Thus, when we prove that a finite group is not simple, we are proving that it has some hidden normal structure. The Sylow theorems are especially useful because they often identify this structure from the order of the group alone. :::theorem Let $G$ be a finite group and let $p$ be a prime divisor of $|G|$. If the number $n_p$ of Sylow $p$-subgroups of $G$ is equal to $1$, then $G$ is not simple whenever the unique Sylow $p$-subgroup is neither $\{e\}$ nor $G$. ::: :::proof Given that $G$ is a finite group, $p$ is a prime divisor of $|G|$, and $n_p=1$. To prove that $G$ is not simple whenever the unique Sylow $p$-subgroup is neither $\{e\}$ nor $G$. Let $P$ be the unique Sylow $p$-subgroup of $G$. For any $g \in G$, the conjugate $gPg^{-1}$ is also a Sylow $p$-subgroup of $G$. Since $P$ is the unique Sylow $p$-subgroup, we have $$\begin{equation*} gPg^{-1}=P \end{equation*}$$ for all $g \in G$. Therefore $P \trianglelefteq G$. If $P \neq \{e\}$ and $P \neq G$, then $P$ is a nontrivial proper normal subgroup of $G$. Hence $G$ is not simple. $\square$ ::: This preview turns the non-simplicity tests into a decision path. Choose a group order from the examples, and the flow identifies the main reason simplicity fails. Abelian nonprime orders fail because every subgroup is normal, p-power orders fail because the center is nontrivial, and many order $pq$ cases fail because the larger-prime Sylow subgroup is unique. The flow helps you decide which theorem to use before doing detailed calculations. :::scientific-preview[Simple Group Test Flowchart] ::: Every path ends by finding a normal subgroup that is neither trivial nor the whole group. That is exactly what contradicts simplicity. This calculator checks the most common Sylow test for non-simplicity. Enter the group order and a prime divisor. The calculator lists the possible values of $n_p$ and decides whether a unique Sylow subgroup is forced. It also checks whether the Sylow subgroup is proper, because a normal Sylow subgroup proves non-simplicity only when it is nontrivial and proper. :::calculator[Sylow Non-Simplicity Test Calculator] ::: The calculator gives a valid non-simplicity proof only when it finds a unique, proper, nontrivial Sylow subgroup. If multiple Sylow counts remain possible, another theorem or argument is needed. :::theorem Let $G$ be a finite abelian group. Then $G$ is simple if and only if $G$ has prime order. ::: :::proof Given that $G$ is a finite abelian group. To prove that $G$ is simple if and only if $G$ has prime order. First suppose that $G$ has prime order $p$. By Lagrange's theorem, the only subgroups of $G$ have orders $1$ and $p$. Hence the only subgroups of $G$ are $\{e\}$ and $G$. Therefore the only normal subgroups of $G$ are $\{e\}$ and $G$. Hence $G$ is simple. Conversely, suppose that $G$ is simple. If $|G|$ is not prime, then $|G|$ has a prime divisor $p$ with $1<p<|G|$. By Cauchy's theorem, $G$ has an element $a$ of order $p$. Hence $\langle a\rangle$ is a nontrivial proper subgroup of $G$. Since $G$ is abelian, $\langle a\rangle$ is normal in $G$, contradicting simplicity. Therefore $|G|$ is prime. Hence $G$ is simple if and only if $G$ has prime order. $\square$ ::: :::example Let $G$ be a group of order $10$. Then $G$ is not simple. ::: :::solution Let $|G|=10=2\cdot 5$. Let $n_5$ be the number of Sylow $5$-subgroups of $G$. By Sylow's third theorem, $$\begin{align*} n_5 &\equiv 1 \pmod 5, \\ n_5 &\mid 2. \end{align*}$$ The positive divisors of $2$ are $1$ and $2$. Among these, only $1$ is congruent to $1$ modulo $5$. Therefore $$\begin{equation*} n_5=1. \end{equation*}$$ Thus $G$ has a unique Sylow $5$-subgroup $P$. Since $|P|=5$, we have $$\begin{equation*} 1<|P|<10. \end{equation*}$$ The unique Sylow subgroup $P$ is normal in $G$. Hence $G$ has a nontrivial proper normal subgroup, so $G$ is not simple. $\square$ ::: :::example Let $G$ be a group of order $9$. Then $G$ is not simple. ::: :::solution Let $|G|=9=3^2$. Then $G$ is a finite $3$-group. A finite $p$-group has nontrivial centre, so $$\begin{equation*} Z(G)\neq \{e\}. \end{equation*}$$ If $Z(G)\neq G$, then $Z(G)$ is a nontrivial proper normal subgroup of $G$. Hence $G$ is not simple. If $Z(G)=G$, then $G$ is abelian. Since $|G|=9$ is not prime, an abelian group of order $9$ is not simple. Therefore $G$ is not simple in all cases. $\square$ ::: :::theorem Let $p$ be a prime integer and let $n>1$. Then no group of order $p^n$ is simple. ::: :::proof Given that $p$ is a prime integer, $n>1$, and $G$ is a group of order $p^n$. To prove that $G$ is not simple. Since $G$ is a finite $p$-group, its centre is nontrivial: $$\begin{equation*} Z(G)\neq \{e\}. \end{equation*}$$ If $Z(G)\neq G$, then $Z(G)$ is a nontrivial proper normal subgroup of $G$. Therefore $G$ is not simple. If $Z(G)=G$, then $G$ is abelian. Since $|G|=p^n$ with $n>1$, Cauchy's theorem gives a subgroup of order $p$. This subgroup is nontrivial and proper. Since $G$ is abelian, it is normal in $G$. Hence $G$ is not simple. $\square$ ::: :::theorem Let $p$ and $q$ be prime integers. Then no group of order $pq$ is simple. ::: :::proof Given that $p$ and $q$ are prime integers and $G$ is a group of order $pq$. To prove that $G$ is not simple. If $p=q$, then $|G|=p^2$, and the result follows from the non-simplicity of groups of order $p^n$ for $n>1$. Now suppose that $p\neq q$. Without loss of generality, let $p>q$. Let $n_p$ be the number of Sylow $p$-subgroups of $G$. By Sylow's third theorem, $$\begin{align*} n_p &\equiv 1 \pmod p, \\ n_p &\mid q. \end{align*}$$ Thus $n_p=1$ or $n_p=q$. Since $q<p$, the congruence $q\equiv 1 \pmod p$ is impossible. Therefore $$\begin{equation*} n_p=1. \end{equation*}$$ Hence the Sylow $p$-subgroup is unique, normal, nontrivial, and proper. Hence $G$ is not simple. $\square$ ::: :::solved-problem Let $G$ be a group of order $21$. Prove that $G$ is not simple. ::: :::solution Let $|G|=21=3\cdot 7$. Let $n_7$ be the number of Sylow $7$-subgroups of $G$. By Sylow's third theorem, $$\begin{align*} n_7 &\equiv 1 \pmod 7, \\ n_7 &\mid 3. \end{align*}$$ The positive divisors of $3$ are $1$ and $3$. Since $3\not\equiv 1 \pmod 7$, we get $$\begin{equation*} n_7=1. \end{equation*}$$ Thus the Sylow $7$-subgroup is unique and normal. It has order $7$, so it is neither $\{e\}$ nor $G$. Hence $G$ is not simple. $\square$ ::: :::exercise 1. Prove that no group of order $14$ is simple. 2. Prove that no group of order $25$ is simple. 3. Let $G$ be a group of order $33$. Show that $G$ has a normal Sylow $11$-subgroup. ::: :::answer 1. If $|G|=14=2\cdot 7$, then $n_7\equiv 1 \pmod 7$ and $n_7\mid 2$, so $n_7=1$. Thus the Sylow $7$-subgroup is normal and proper. 2. A group of order $25=5^2$ is a finite $5$-group with nontrivial centre. If its centre is proper, it is a nontrivial proper normal subgroup; if the centre is the whole group, the group is abelian of nonprime order and is not simple. 3. If $|G|=33=3\cdot 11$, then $n_{11}\equiv 1 \pmod {11}$ and $n_{11}\mid 3$, so $n_{11}=1$. ::: :::faq[Frequently Asked Questions] Q: Why does uniqueness of a Sylow subgroup imply normality? A: Conjugates of a Sylow $p$-subgroup are again Sylow $p$-subgroups. If there is only one such subgroup, every conjugate must equal it. Q: Does every Sylow subgroup prove non-simplicity? A: No. A Sylow subgroup proves non-simplicity only when it is normal, nontrivial, and proper. Q: Why are abelian simple groups so restricted? A: In an abelian group every subgroup is normal, so a finite abelian simple group can have no nontrivial proper subgroup. This forces prime order. :::