## Review of Group Theory: Group Actions (Pt. IV Conjugation and the Class Equation Pt. II)

**Point of post: **This is a continuation of this post.

From this we can derive some interesting results, but first we need some notation. For any group (not necessarily a -group) and any we call the set

(note that need only be a subset) the *normalizer *of and denote this . It’s clear that is a subgroup of and if itself is group it’s the largest (by set containment) normal group containing . With this in mind we have:

**Theorem: ***Let be a group such that where is prime and . Then:*

* *

* *

**Proof:**

: Let be a set of one representative from each conjugacy class of with more than one element. By the class equation we have that

note though that for each we have that so that . Thus,

and thus consequently

and since it follows that for some . The conclusion follows.

: Recall from our proof the class equation that

and thus

and thus

notice though that if then

(since ) and if then . It follows then that

but using the same logic as before, noting that , we may conclude that and thus is non-trivial.

: This last result follows by noticing that which is the smallest prime dividing and the conclusion follows from an earlier result.

**Corollary: ***Every group of order where is a prime is abelian.*

**Proof: **By the above theorem we have that is non-trivial and so by Lagrange’s Theorem we have that . Assume that then and thus (from previous theorem) we may conclude that is cyclic. But, from an earlier theorem we may then conclude that is abelian and thus which is a contradiction. Thus, it follows that and thus from where the conclusion follows.

**References:**

1. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. *Basic Abstract Algebra*. Cambridge: Cambridge UP, 1994. Print.

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

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

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