## Review of Group Theory: Consequences of Lagrange’s Theorem

**Point of post: **In this post we give a few interesting theorems which are near-corollaries of Lagrange’s Theorem.

*Motivation*

We stated in our last post on Lagrange’s theorem that it was powerful. Indeed, it itself has wonderful, wonderful properties. But, we can use it to prove other, beautiful theorems. That is what this post is about.

This first one, although very simple shall be one of the most useful in our next post.

**Theorem: ***Let be a finite group and . Then, .*

**Proof: **We know from Lagrange’s theorem that divides and thus there exists some such that . Thus,

**Theorem: ***Let be a non-trivial group. Then, has no non-trivial proper subgroups if and only if is finite and of order where is prime. Moreover, every group of order is prime is cyclic.*

**Proof:** First assume that has no non-trivial proper subgroups. We claim that is cyclic. To do this we choose with .We note then that and thus by assumption we must have that . We recall though this implies that or for some . Clearly we can’t have that otherwise there would exist some isomorphism and which is a contradiction. Thus, is finite. Now let be such that (). But, we know that every cyclic group has a subgroup of order where divides the order of the ambient group. In particular for every such that has a subgroup of order . It follows by assumption that implies that and thus is prime.

Conversely, if |=p is prime then every subgroup must have order which divides and thus the only subgroups are of order from where it follows that has no non-trivial proper subgroups. But, from the first sentence of the previous previous paragraph we may also conclude that is cyclic.

**Corollary: ***Let be a group with with prime. Then, every proper subgroup of is cyclic.*

Another interesting theorem which one can use Lagrange’s theorem to prove is the following

**Theorem: ***Let be a finite group and . Then,*

*where is the product set defined in our last post.*

**Proof: **Note first that

but since for each it suffices to the number of elements of of the form . To do this we note that if and only if which is true if and only if which is true if and only if . Thus, it follows that the distinct number of elements of of the form is equal to . But, by Lagrange’s theorem we may then conclude that

Along similar lines

**Theorem: ***Let be a finite group and be such that . Then, is trivial.*

**Proof: **We merely note that

and

so that divides both and . It follows that which implies that is trivial.

Lastly

**Theorem: ***Let be a finite group and be such that where is prime. Then, if then or .*

**Proof: **We note by of Lagrange’s theorem that

from where it follows that either or . But, noting that for any subgroup of we have that it follows that if then . From this the conclusion follows.

**References:**

1. Lang, Serge. *Undergraduate Algebra*. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 200

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