Let be a group. A subgroup is called maximal if and the only subgroups of which contain are and .
- Prove that if is a proper subgroup of the finite group then there exists a maximal subgroup of which contains .
- Prove that is maximal.
- Show that if is a cyclic group of order then a subgroup is maximal if and only if for some prime dividing .
- Consider the set . This is a subset of and so is partially ordered by . Moreover, is finite. If , then itself is maximal. If is not empty and we suppose that no maximal subset containing exists, then there is an element . Since is not maximal, there exists an element with . Similarly, exists in with , and we may construct an infinite ascending chain of subgroups . This is a contradiction since is finite. Thus a maximal subgroup containing must exist.
- We know that . Now if , then by Lagrange's Theorem. Thus is maximal.
- We first prove a simple lemma.
Lemma 1: Let be a finite cyclic group of order and and be subgroups of . Then if and only if for some , mod . Proof: We have , so that for some integer . We saw in a previous exercise that this implies mod . If mod , in . So so that .
Now for the main result.
Suppose is cyclic of order , and let be a subgroup of , and let be minimal such that generates .
Suppose is maximal. Say where and are relatively prime. By the lemma, and . Since is maximal, we have . Suppose that ; then we have by the minimalness of . In particular, . But then , and so , a contradiction. Thus either or is 1, and so is prime. Now by Theorem 7 in D&F, and again by the minimalness of , we have that divides .
Suppose is a prime dividing . Suppose further that we have a subgroup such that . By the lemma, we have . Thus either , so that , or , so that . Thus is maximal.