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.

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