While studying for the algebra qual, I stumbled upon the following problem:

Let be a group of order where is an odd prime. Prove that contains a nontrivial normal subgroup.

The problem is simple enough, however, it uses many results from group theory. We’ll investigate the problem in detail bellow.

**Cosests and Lagrange’s Theorem**

**Definition. **Let be a subgroup of a group . The subset of is the *left coset* of containing . The subset is the* right coset* of containing .

Cosets are important because they allow us to define normal subgroups, which in turn allow us to construct factor groups. We will need two important results concerning cosets.

If , then the set of left cosets of form a partition of . Indeed, let the relation be defined on by if and only if . It is easily verified that is an equivalence relation on with equivalence class . Moreover, if and only if . This result follows a simple set containment argument along with the definition of left coset. It should be noted that both of these results have an analogue for right cosets.

**Lagrange’s Theorem. **If is a finite group and is a subgroup of , then the order of divides the order of .

*Proof. *Let and . I claim that every left coset of has order . Define the function by . Then is surjective by the definition of as . Suppose that for and in . Then and by the cancellation law of a group, . Thus, is injective. Since is a bijection, we conclude that and have the same order and the claim is established. Since the left cosets of form a partition of , we have that

.

Thus,

.

Since the order of each left coset of is equal to , we have that

where is summed up times. That is, which shows that is a divisor of .

In our proof, to establish that every left coseet of has the same order as we showed that the function was a bijection. It turns out that every right coset of also has the same order as . To establish this, check that defined by is a bijection.

Lagrange’s theorem is perhaps one of the most useful results in group theory since it allows us to actually compute orders of subgroups. The converse of Lagrange’s theorem is not true. The classic counter example is the alternating group , which is the group of even permutations on elements. The order of is and is a divisor of , but has no subgroup of order . There are however some partial converses to Lagrange’s theorem. A partial converse which holds for arbitrary finite groups is the following result.

**Cauchy’s Theorem.** If is a finite group and is a prime dividing , then has an element of order .

Cauchy’s theorem will be useful for us in proving our original problem.

**Normal Subgroups**

**Definition. **A subgroup of a group is *normal* if for all . If is a normal subgroup of then we shall write .

The next theorem is of interest to us since it will be essential for proving our original problem.

**Theorem 1. **Let be a group of order . If is a subgroup of and , then .

*Proof. *Let . Then , , and . Since , we have that and . The elements of cosets are* *distinct because they from a partition of , thus, it must be the case that .

**The Original Problem**

Let be a group of order where is an odd prime. Prove that contains a nontrivial normal subgroup.

*Proof.* By Cauchy’s theorem, there exists an element such that . Then the cyclic subgroup is nontrivial and has order . By theorem 1, is a normal subgroup of .

The proof is quite elegant, but as you have seen, we needed to recall some important results in order achieve such a simple proof.

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