The index multiplication formula has a sharp consequence when the index is prime. If a subgroup $H$ has prime index in a finite group $G$, then no subgroup can sit strictly between $H$ and $G$. This is a structural application of Lagrange's theorem: the index from $H$ to $G$ cannot split into two nontrivial positive factors. In this lecture, we prove the result and use it to identify when a subgroup is maximal with respect to inclusion. :::definition[Subgroup of Prime Index] Let $(G,\circ)$ be a finite group and let $H$ be a subgroup of $G$. The subgroup $H$ is called a $subgroup of prime index$ if $$\begin{equation*} [G:H]=p \end{equation*}$$ for some prime number $p$. ::: :::definition[Maximal Subgroup] Let $(G,\circ)$ be a group and let $H$ be a proper subgroup of $G$. The subgroup $H$ is called a $maximal subgroup$ of $G$ if there is no subgroup $K$ of $G$ such that $$\begin{equation*} H<K<G. \end{equation*}$$ ::: :::theorem Let $(G,\circ)$ be a finite group and let $H$ be a subgroup of $G$. If $[G:H]$ is prime, then $H$ is a maximal subgroup of $G$. ::: :::proof Given that $(G,\circ)$ is a finite group, $H$ is a subgroup of $G$, and $[G:H]$ is prime. To prove that $H$ is a maximal subgroup of $G$. Let $$\begin{equation*} [G:H]=p, \end{equation*}$$ where $p$ is prime. Let $K$ be a subgroup of $G$ such that $$\begin{equation*} H\leq K\leq G. \end{equation*}$$ By the index multiplication formula, $$\begin{equation*} [G:H]=[G:K][K:H]. \end{equation*}$$ Therefore $$\begin{equation*} p=[G:K][K:H]. \end{equation*}$$ Since $p$ is prime, either $$\begin{equation*} [G:K]=1 \end{equation*}$$ or $$\begin{equation*} [K:H]=1. \end{equation*}$$ If $[G:K]=1$, then $K=G$. If $[K:H]=1$, then $K=H$. Therefore every subgroup $K$ satisfying $H\leq K\leq G$ is equal to $H$ or $G$. Hence, $H$ is a maximal subgroup of $G$. $\square$ ::: We observe the interval of possible subgroups between $H$ and $G$ through the factorization of $[G:H]$. Enter an index and compare the available factorizations of that index. When the index is prime, there is no way to split it into two factors both larger than $1$, so an intermediate subgroup cannot appear through the index multiplication formula. When the index is composite, the theorem does not decide maximality, but the factorization shows why the prime argument no longer works. :::scientific-preview[Prime Index Maximality Explorer] ::: The absence of nontrivial factorization is the key point. Prime index forces every subgroup between $H$ and $G$ to collapse to one endpoint, which is exactly maximality. :::corollary Let $(G,\circ)$ be a finite group and let $H$ be a subgroup of $G$. If $$\begin{equation*} |G|=p|H| \end{equation*}$$ for some prime number $p$, then $H$ is a maximal subgroup of $G$. ::: :::proof Given that $(G,\circ)$ is a finite group, $H$ is a subgroup of $G$, and $|G|=p|H|$ for some prime number $p$. To prove that $H$ is a maximal subgroup of $G$. By Lagrange's theorem, $$\begin{align*} [G:H] &=\frac{|G|}{|H|}\\ &=\frac{p|H|}{|H|}\\ &=p. \end{align*}$$ Since $p$ is prime, $[G:H]$ is prime. By the previous theorem, $H$ is a maximal subgroup of $G$. Hence, $H$ is a maximal subgroup of $G$. $\square$ ::: :::example[A Maximal Subgroup in Integers Modulo Six] Let $G=\mathbb{Z}_6$ under addition modulo $6$ and let $$\begin{equation*} H=\{0,3\}. \end{equation*}$$ Then $$\begin{equation*} |G|=6 \end{equation*}$$ and $$\begin{equation*} |H|=2. \end{equation*}$$ Therefore $$\begin{align*} [G:H] &=\frac{6}{2}\\ &=3. \end{align*}$$ Since $3$ is prime, $H$ is a maximal subgroup of $\mathbb{Z}_6$. ::: :::example[Prime Index Does Not Mean Prime Order] Let $G=\mathbb{Z}_{12}$ and let $$\begin{equation*} H=\{0,3,6,9\}. \end{equation*}$$ Then $|H|=4$ and $$\begin{equation*} [G:H]=\frac{12}{4}=3. \end{equation*}$$ Thus $H$ has prime index, but $H$ does not have prime order. Prime index is a statement about the number of cosets, not about the number of elements inside the subgroup. ::: :::solved-problem[Testing Maximality by Index] Let $(G,\circ)$ be a finite group with $|G|=30$. Let $H$ be a subgroup of $G$ with $|H|=10$. Prove that $H$ is maximal. ::: :::solution[Solution] Let $(G,\circ)$ be a finite group with $|G|=30$ and let $H$ be a subgroup of $G$ with $|H|=10$. By Lagrange's theorem, $$\begin{align*} [G:H] &=\frac{|G|}{|H|}\\ &=\frac{30}{10}\\ &=3. \end{align*}$$ Since $3$ is prime, $H$ is a maximal subgroup of $G$. ::: We compute the index $[G:H]$ from the orders of $G$ and $H$. If the index is prime, the theorem proves that $H$ is maximal in $G$. If the index is not prime, the calculator says only that this theorem does not apply. This avoids the common mistake of treating a failed sufficient condition as proof that $H$ is not maximal. :::calculator[Prime Index Maximality Calculator] ::: :::exercise[Exercises] 1. Let $G$ be a finite group with $|G|=42$ and let $H$ be a subgroup with $|H|=14$. Prove that $H$ is maximal. 2. Let $G$ be a finite group with $|G|=48$ and let $H$ be a subgroup with $|H|=12$. Does the prime-index theorem prove that $H$ is maximal? 3. Let $H\leq K\leq G$ and suppose $[G:H]=5$. Prove that $K=H$ or $K=G$. ::: :::answer[Answers] 1. We have $$\begin{equation*} [G:H]=\frac{42}{14}=3. \end{equation*}$$ Since $3$ is prime, $H$ is maximal. 2. We have $$\begin{equation*} [G:H]=\frac{48}{12}=4. \end{equation*}$$ Since $4$ is not prime, the prime-index theorem does not apply. This does not decide maximality. 3. By the index multiplication formula, $$\begin{equation*} [G:H]=[G:K][K:H]. \end{equation*}$$ Thus $$\begin{equation*} 5=[G:K][K:H]. \end{equation*}$$ Since $5$ is prime, either $[G:K]=1$ or $[K:H]=1$. Hence, $K=G$ or $K=H$. ::: :::faq Q: What does prime index mean? A: It means the number of cosets of $H$ in $G$ is a prime number. Q: Why does prime index imply maximality? A: An intermediate subgroup would split the prime index into two positive factors, which is impossible except by using $1$. Q: Does maximality imply prime index? A: Not in general. The theorem in this lecture gives one sufficient condition for maximality. ::: :::call-to-action subtitle: Continue with double cosets, where two subgroups act on both sides of a group element. button: Next Section: Double Cosets | published/abstract-algebra/subgroups-and-normal-subgroups/double-cosets/double-coset-definition.md :::