## Review of Group Theory: Group Actions (Pt. I Definitions and a Sharpening of Cayley’s Theorem cont.)

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

*A Sharpening of Cayley’s Theorem*

With all these preliminary definitions out of the way we are now able to state intelligently a sharpening of Cayley’s theorem which will give us a vast generalization of the commonly used theorem that if is a group and with then .

We begin by noticing that we have already encountered a very nice example of a group action. In particular, for any group and any we have that acts on (as a set, it isn’t necessarily a group since we haven’t assumed that ) by . This is evidently a group actions since for any and any we have that

and

We call this action *left multiplication on the cosets of . W*ith this observation we are now able to formulate the following theorem:

**Theorem: ***Let be a group and . Let be the action of left multiplication on the cosets of and let . Then, and*

*where is the inner automorphism induced by .*

**Proof: **By definition we have that

Also,

$latex\ text{ }$

It may seem unobvious at first how this is a sharpening of Cayley’s theorem, but if one takes in the above theorem then one can naturally identify with and thus will be a homomorphism into whose kernel is (since and so all the “terms” in the above intersection will be trivial) and so will be isomorphic to its image.

We end this first post with a generalization of the common theorem that a subgroup of index is normal. Namely:

**Theorem: ***Let be finite group and with where is the smallest prime which divides . Then, .*

**Proof: **Let act on by multiplication on the cosets of and let be the associated homomorphism and . From Lagrange’s Theorem we know that

where . By the First Isomorphism Theorem we have that . But, by Lagrange’s Theorem this implies that and thus . Suppose then that is a prime and , then but since every prime divisor of is strictly less then it follows that and thus since we have that . But, this contradicts the minimality of and thus there are no such . But, this clearly implies (since the only natural number with no prime divisors is ) that . Recalling though that this implies that and thus as required.

**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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