Cauchy's theorem gives prime-order subgroups, but normality requires more structure. The focus keyword is normal subgroups from Cauchy's theorem, because many standard arguments begin with an element of prime order and then use group actions, indices, uniqueness, and conjugation to prove normality. These methods prepare students for Sylow's normality criteria. In this lesson, the aim is not to prove every subgroup normal, but to recognize the situations where prime-order existence forces a normal subgroup. :::definition[Normal Subgroup] Let $G$ be a group and let $N\leq G$. Then $N$ is called a normal subgroup of $G$ if $$\begin{equation*} gNg^{-1}=N \end{equation*}$$ for every $g\in G$. We write $N\trianglelefteq G$. ::: :::definition[Conjugation Action on Subgroups] Let $G$ be a group and let $H\leq G$. The conjugation action on subgroups sends an element $g\in G$ to the subgroup $$\begin{equation*} gHg^{-1}. \end{equation*}$$ A subgroup is normal exactly when its conjugacy orbit under this action has one element. ::: :::lemma Let $G$ be a finite group and let $H\leq G$. If $H$ is the unique subgroup of $G$ of its order, then $H$ is normal in $G$. ::: :::proof Given that $G$ is a finite group and $H$ is the unique subgroup of $G$ of order $|H|$. To prove that $H$ is normal in $G$. Let $g\in G$. Then $gHg^{-1}$ is a subgroup of $G$, and conjugation is a bijection from $H$ to $gHg^{-1}$. Therefore $|gHg^{-1}|=|H|$. By uniqueness, $gHg^{-1}=H$. Since this holds for every $g\in G$, $H$ is normal in $G$. Hence $H\trianglelefteq G$. $\square$ ::: :::theorem Let $G$ be a group of order $pq$, where $p$ and $q$ are primes with $p<q$. If $p$ does not divide $q-1$, then $G$ has a normal subgroup of order $q$. ::: :::proof Given that $|G|=pq$, where $p$ and $q$ are primes with $p<q$, and $p$ does not divide $q-1$. To prove that $G$ has a normal subgroup of order $q$. By Cauchy's theorem, $G$ has a subgroup $Q$ of order $q$. Conjugation by elements of $G$ gives a homomorphism $$\begin{equation*} \theta:G\to\operatorname{Aut}(Q). \end{equation*}$$ Since $Q$ has prime order $q$, it is cyclic, and $|\operatorname{Aut}(Q)|=q-1$. The image $\theta(G)$ has order dividing both $|G|=pq$ and $q-1$. Since $q$ does not divide $q-1$ and $p$ does not divide $q-1$, the only possible order is $1$. Therefore conjugation by every element of $G$ fixes $Q$ setwise. Thus $gQg^{-1}=Q$ for every $g\in G$. Hence $Q\trianglelefteq G$, and $G$ has a normal subgroup of order $q$. $\square$ ::: :::theorem Let $G$ be a finite group and let $H\leq G$ have index $2$. Then $H$ is normal in $G$. ::: :::proof Given that $G$ is a finite group and $H\leq G$ with $[G:H]=2$. To prove that $H$ is normal in $G$. The subgroup $H$ has exactly two left cosets and exactly two right cosets. One left coset is $H$, and the other is $G\setminus H$. One right coset is $H$, and the other is $G\setminus H$. Therefore for every $g\in G$, $gH=Hg$. Thus $gHg^{-1}=H$ for every $g\in G$. Hence $H\trianglelefteq G$. $\square$ ::: :::example Let $G$ be a group of order $15$. Since $15=3\cdot5$ and $3$ does not divide $5-1=4$, the theorem on groups of order $pq$ applies with $p=3$ and $q=5$. Therefore $G$ has a normal subgroup of order $5$. ::: :::example Let $G$ be a group of order $14$. Since $14=2\cdot7$ and $2$ divides $7-1=6$, the preceding theorem does not apply. However, Cauchy's theorem still guarantees a subgroup of order $7$ and a subgroup of order $2$. ::: This preview compares three common normality mechanisms. Cauchy's theorem gives prime-order subgroups, but normality may come from uniqueness, index $2$, or the $pq$ criterion. Choose an example and read which mechanism applies. The examples $15$, $14$, $21$, and $28$ show that the existence of a subgroup and the normality of that subgroup are different claims. :::scientific-preview[Prime Order Normality Decision Explorer] ::: Normality is a statement about how the whole group conjugates a subgroup. Cauchy's theorem supplies prime-order subgroups, but extra conditions decide whether those subgroups are fixed under conjugation. This calculator checks the theorem for groups of order $pq$ with $p<q$. Enter two primes, and the calculator determines whether $p$ divides $q-1$. If it does not, a normal subgroup of order $q$ is guaranteed. The calculator also includes an index-two check because that is another common normality shortcut in this lesson. :::calculator[pq Normal Subgroup Test Calculator] ::: The calculator reports when a theorem applies, not when normality is impossible. A failed criterion means the given shortcut does not prove normality; another argument may still work. :::solved-problem Prove that every group of order $15$ has a normal subgroup of order $5$. ::: :::solution Given that $|G|=15=3\cdot5$. Here $p=3$ and $q=5$, with $p<q$. Since $3\nmid4$, the theorem on groups of order $pq$ applies. Therefore $G$ has a normal subgroup of order $5$. Hence every group of order $15$ has a normal subgroup of order $5$. ::: :::solved-problem Let $G$ be a group of order $28$. Prove that any subgroup of order $14$ is normal. ::: :::solution Given that $G$ is a group of order $28$ and $H\leq G$ with $|H|=14$. Then $$\begin{align*} [G:H]&=\frac{|G|}{|H|}\\ &=\frac{28}{14}\\ &=2. \end{align*}$$ Every subgroup of index $2$ is normal. Therefore $H\trianglelefteq G$. Hence any subgroup of order $14$ is normal. ::: :::exercise 1. Prove that every subgroup of index $2$ in a finite group is normal. 2. Let $|G|=21$. Use the theorem on groups of order $pq$ to decide whether a normal subgroup of order $7$ is guaranteed. 3. Let $H$ be the unique subgroup of order $p$ in $G$. Prove that $H$ is normal. 4. Explain why Cauchy's theorem alone does not imply normality. ::: :::answer 1. A subgroup of index $2$ has two left cosets and two right cosets; the complement of $H$ is the only nonidentity coset on both sides, so left and right cosets agree. 2. Since $21=3\cdot7$ and $3$ divides $7-1=6$, the stated theorem does not guarantee normality. 3. For any $g\in G$, the subgroup $gHg^{-1}$ has order $p$. By uniqueness, $gHg^{-1}=H$, so $H$ is normal. 4. Cauchy's theorem gives existence of a subgroup of prime order; it does not say that conjugation preserves that subgroup. ::: :::faq[Frequently Asked Questions] Q: Does every subgroup of prime order have to be normal? A: No. Prime order controls the subgroup internally, not how the whole group conjugates it. Q: Why does uniqueness imply normality? A: Conjugation preserves subgroup order. If there is only one subgroup of that order, every conjugate must be the same subgroup. Q: Why are groups of order $pq$ important? A: They are the first setting where prime divisors, Cauchy's theorem, and normality tests interact in a meaningful way. ::: :::call-to-action[Compute p-Subgroups Explicitly] subtitle: Turn the theory into concrete subgroup calculations in cyclic and permutation groups. button: Computing p-Subgroups | /abstract-algebra/sylow-theorems/cauchys-theorem-and-p-groups/computing-p-subgroups button-ghost: Subgroups of Finite p-Groups | /abstract-algebra/sylow-theorems/cauchys-theorem-and-p-groups/subgroups-of-finite-p-groups :::