## Characterization of maximal subgroups of a finite cyclic group

Let $G$ be a group. A subgroup $M \leq G$ is called maximal if $M \neq G$ and the only subgroups of $G$ which contain $M$ are $M$ and $G$.

1. Prove that if $H$ is a proper subgroup of the finite group $G$ then there exists a maximal subgroup of $G$ which contains $H$.
2. Prove that $\langle r \rangle \leq D_{2n}$ is maximal.
3. Show that if $G = \langle x \rangle$ is a cyclic group of order $n \geq 1$ then a subgroup $H \leq G$ is maximal if and only if $H = \langle x^p \rangle$ for some prime $p$ dividing $n$.

1. Consider the set $S = \{ K < G \ |\ H \leq K \}$. This is a subset of $\mathcal{P}(G)$ and so is partially ordered by $\subseteq$. Moreover, $S$ is finite. If $S = \emptyset$, then $H$ itself is maximal. If $S$ is not empty and we suppose that no maximal subset containing $H$ exists, then there is an element $K_0 \in S$. Since $K_0$ is not maximal, there exists an element $K_1 \in S$ with $K_0 < K_1$. Similarly, $K_2$ exists in $S$ with $K_1 < K_2$, and we may construct an infinite ascending chain of subgroups $K_0 < K_1 < K_2 < \cdots$. This is a contradiction since $S$ is finite. Thus a maximal subgroup containing $H$ must exist.
2. We know that $|\langle r \rangle| = n$. Now if $\langle r \rangle < A \leq D_{2n}$, then $A = D_{2n}$ by Lagrange's Theorem. Thus $\langle r \rangle$ is maximal.
3. We first prove a simple lemma.

Lemma 1: Let $G = \langle x \rangle$ be a finite cyclic group of order $n$ and $H = \langle x^a \rangle$ and $K = \langle x^b \rangle$ be subgroups of $G$. Then $H \leq K$ if and only if for some $k$, $a = kb$ mod $n$. Proof: $(\Rightarrow)$ We have $x^a \in \langle x^b \rangle$, so that $x^a = x^{bk}$ for some integer $n$. We saw in a previous exercise that this implies $a = bk$ mod $n$. $(\Leftarrow)$ If $a = bk$ mod $n$, $x^a = x^{bk}$ in $G$. So $x^a \in \langle x^b \rangle$ so that $H \leq K$. $\square$

Now for the main result.

Suppose $G = \langle x \rangle$ is cyclic of order $n$, and let $H = \langle x^k \rangle$ be a subgroup of $G$, and let $k$ be minimal such that $x^k$ generates $H$.

$(\Rightarrow)$ Suppose $H \leq G$ is maximal. Say $k = ab$ where $a$ and $b$ are relatively prime. By the lemma, $H \leq \langle x^a \rangle$ and $H \leq \langle x^b \rangle$. Since $H$ is maximal, we have $\langle x^a \rangle, \langle x^b \rangle \in \{H, G\}$. Suppose that $a,b < k$; then we have $\langle x^a \rangle = \langle x^b \rangle = G$ by the minimalness of $k$. In particular, $\mathsf{gcd}(a,n) = \mathsf{gcd}(b,n) = 1$. But then $\mathsf{gcd}(k,n) = \mathsf{gcd}(ab,n) = 1$, and so $H = G$, a contradiction. Thus either $a$ or $b$ is 1, and so $k$ is prime. Now $\langle x^k \rangle = \langle x^{\mathsf{gcd}(k,n)} \rangle$ by Theorem 7 in D&F, and again by the minimalness of $k$, we have that $k = \mathsf{gcd}(k,n)$ divides $n$.

$(\Leftarrow)$ Suppose $k$ is a prime dividing $n$. Suppose further that we have a subgroup $K = \langle x^m \rangle$ such that $H \leq K \leq G$. By the lemma, we have $m | k$. Thus either $m = 1$, so that $K = G$, or $m = k$, so that $K = H$. Thus $H$ is maximal.