Exercise 2.4.16 (Maximal subgroups of $G$)

Proof. Let $G$ be a finite group.

(a)

Let $H$ be a proper subgroup of $G$. Consider the subset $A$ of $=\{0,1,2,3,\dots \}$ defined by $$A=\{n\in \mid \exists <!--\; FUNCTION\; APPLICATION\; -->K,n=|K|\text{ and }H\le K<G\}.$$

( $K<G$ means $K\le G$ and $K\ne G$.)

Since $H$ is a proper subgroup of $G$, $H$ satisfies $H\le H<G$, so $n_0=|H|\in A$. Therefore $A\ne \varnothing$.

Moreover, every $K$ such that $H\le K<G$ satisfies $|K|<|G|$, so $A$ is upper bounded by $|G|$.

Hence $A$ has a greatest element $a=\max <!--\; FUNCTION\; APPLICATION\; -->(A)$. Since $a\in A$, there exists a subgroup $M$ such that

$$|M|=\max <!--\; FUNCTION\; APPLICATION\; -->(A),H\le M<G.$$

So $M\ne G$, and if $K$ is a subgroup of $G$ such that $M\le K<G$, then $|K|\in A$, thus $|K|\le \max <!--\; FUNCTION\; APPLICATION\; -->(A)=|M|$. Then $M\subseteq K$ and $|M|=|K|$, where $K$ is finite, therefore $M=K$. This shows that $M$ is a maximal subgroup of $G$ containing $H$.

If $H$ is a proper subgroup of the finite group $G$ then there is a maximal subgroup of $G$ containing $H$.

(b)

Consider the dihedral group $D_{2n}=⟨r,s\mid r^n=s^2=1,\mathit{sr}=r{s}^{-1}⟩$ and the subgroup $H$ of all rotations, i.e., $$H=⟨r⟩=\{1,r,r^2,\cdots ,r^{n-1}\}.$$

We know that $|H|=n$ (thus $H\ne D_{2n}$). If $H\le K<D_{2n}$, then $n\le |K|<2n$, and by Lagrange’s Theorem, $|K|$ is a divisor of $2n$. The only divisor $d$ of $2n$ such that $n\le d<2n$ is $d=n$, therefore $|K|=n=|H|$. Since $H\subseteq K$, this proves $H=K$, so $H$ is a maximal subgroup of $D_{2n}$.

(c)

Suppose that $H=⟨x^p⟩$ for some prime $p$ dividing $n=|x|=|G|$. Then $$|H|=|x^p|=\frac{n}{p}.$$

Suppose that $H\le K<G$. Put $d=|K|$. Then $d<n$. The Lagrange’s Theorem shows that

$$\frac{n}{p}\mid d\text{and}d\mid n,(d\ne n).$$

Therefore $\frac{n}{d}\mid p$, where $\frac{n}{d}$ is a positive integer greater than $1$. Since $p$ is a prime number, $\frac{n}{d}=p$. Therefore $|H|=\frac{n}{p}=d=|K|$, where $H\subseteq K$, thus $H=K$. This shows that $H=⟨x^p⟩$ is a maximal subgroup of $G$.

Conversely, suppose that $H$ is a maximal subgroup of $G=⟨x⟩$, where $|G|=n$. By Theorem 7, $H$ is cyclic and $H=⟨x^p⟩$ for some divisor $p$ of $n$. It remains to show that $p$ is a prime number.

First $p\ne 1$, otherwise $H=G$: this is impossible because $H$ is a maximal subgroup of $G$, so $H\ne G$ by definition.

If $a>1$ is a divisor of $p$, then $1<a\le p$ and there is some integer $b$ such that $p=\mathit{ab}$. Then $x^p={(x^a)}^b$, thus $x^p\in ⟨x^a⟩$, so $H=⟨x^p⟩\le ⟨x^a⟩$. Since $H$ is a maximal subgroup, $⟨x^a⟩=G$ or $⟨x^a⟩=H$.

• If $⟨x^a⟩=G$, then $⟨x^a⟩=⟨x⟩$, so $x\in ⟨x^a⟩$. There is some integer $k$ such that $x={(x^a)}^k$, thus $x^{\mathit{ak}-1}=1$, where $|x|=n$, so $n\mid \mathit{ak}-1$, where $p\mid n$, therefore $p\mid \mathit{ak}-1$. Since $a\mid p$, $a\mid \mathit{ak}-1$, thus $a\mid 1$, where $1<a$: this is impossible.
• If $⟨x^a⟩=H$, then $⟨x^a⟩=⟨x^p⟩$, so $x^a\in ⟨x^p⟩$, i.e., $x^a={(x^p)}^l$ for some integer $l$. Then $x^{\mathit{pl}-a}=1$, thus $n\mid \mathit{pl}-a$, and $p\mid n$, so $p\mid \mathit{pl}-a$, which shows that $p\mid a$. Since $a\mid p$, this gives $a=p$.

The only positive divisors of $p\ne 1$ are $1$ and $p$, so $p$ is a prime number.

In conclusion, if $G=⟨x⟩$ is a cyclic group of order $n$ then a subgroup $H$ is maximal if and only if $H=⟨x^p⟩$ for some prime $p$ dividing $n$.