Abstract Nonsense

Crushing one theorem at a time

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 G is a group and H\leqslant G with \left(G:H\right)=2 then H\unlhd G.

\text{ }

We begin by noticing that we have already encountered a very nice example of a group action. In particular, for any group G and any H\leqslant G we have that G acts on G/H (as a set, it isn’t necessarily a group since we haven’t assumed that H\unlhd G) by g\cdot kH=(gk)H. This is evidently a group actions since for any g,g'\in G and any kH\in G/H we have that

\text{ }

g\cdot\left(g'\cdot kH\right)=g\cdot (g'k)H=(gg'k)H=(gg')\cdot kH

\text{ }

and

\text{ }

e\cdot kH=(ek)H=kH

\text{ }

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

\text{ }

Theorem: Let G be a group and H\leqslant G. Let \cdot be the action of left multiplication on the cosets of H and let \Psi:G\to S_{G/H}:g\mapsto \theta_g. Then, G_{H}=H and

\text{ }

\displaystyle \ker\Psi=\bigcap_{g\in G}i_g\left(H\right)

\text{ }

where i_g is the inner automorphism induced by g.

Proof: By definition we have that

\text{ }

\displaystyle \begin{aligned}G_H &= \left\{g\in G:g\cdot H=H\right\}\\ &=\left\{g\in G:gH=H\right\}\\ &= H\end{aligned}

\text{ }

Also,

\text{ }

\displaystyle \begin{aligned}\ker\Psi &= \left\{g\in G:g\cdot\left(kH\right)=kH\;\;\text{for all }kH\in G/H\right\}\\ &=\left\{g\in G:(gk)H=kH\;\;\text{ for all }kH\in G/H\right\}\\ &= \left\{g\in G:(k^{-1}gk)H =H\right\}\\ &= \left\{g\in G:kgk^{-1}\in H\;\;\text{for all }k\in G\right\}\\ &= \bigcap_{k\in G}k^{-1}Hk\end{aligned}

\text{ }

\blacksquare

$latex\ text{ }$

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

We end this first post with a generalization of the common theorem that a subgroup of index 2 is normal. Namely:

Theorem: Let G be finite group and H\leqslant G with \left(G:H\right)=p where p is the smallest prime which divides |G|. Then, H\unlhd G.

Proof: Let G act on G/H by multiplication on the cosets of H and let \Psi be the associated homomorphism G\to S_p and K=\ker\Psi. From Lagrange’s Theorem we know that

\text{ }

\left(G:K\right)=\left(G:H\right)\left(H:K\right)=pk

\text{ }

where k=\left(H:K\right). By the First Isomorphism Theorem we have that G/K\cong \Psi\left(G\right). But, by Lagrange’s Theorem this implies that \left|G/K\right|=pk\mid p! and thus k\mid (p-1)!. Suppose then that q is a prime and q\mid k, then q\mid (p-1)! but since every prime divisor of (p-1)! is strictly less then p it follows that q<p and thus since |G|=|K|pk we have that q\mid |G|. But, this contradicts the minimality of p and thus there are no such q. But, this clearly implies (since the only natural number with no prime divisors is 1) that k=1. Recalling though that k=\left(H:K\right) this implies that H=K=\ker\Psi and thus H\unlhd G as required. \blacksquare

\text{ }

\text{ }

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

Advertisements

January 4, 2011 - Posted by | Algebra, Group Theory | , , ,

4 Comments »

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

    Pingback by Review of Group Theory: Group Actions (Pt. IV Conjugation and the Class Equation Pt. II) « Abstract Nonsense | January 6, 2011 | Reply

  2. […] Consider . It’s clear that since (appealing to an earlier theorem). But, it’s also clear that if then . Thus, since by definition we may conclude by our […]

    Pingback by Review of Group Theory: Semidirect Products (Pt. III The Dihedral Group) « Abstract Nonsense | January 11, 2011 | Reply

  3. […] appeal to the common fact that a subgroup of index two is always normal (this is a corollary of a more general fact). But, we note that if the maximum power of that divides is that (since is an odd prime) and […]

    Pingback by University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) (August-2003) « Abstract Nonsense | May 1, 2011 | Reply

  4. […] , since by the corollary we have that and so clearly . That said, we already knew this from the theorem that says if the index of a subgroup is the smallest prime dividing the group then it’s […]

    Pingback by Actions by p-Groups (Pt. II) « Abstract Nonsense | September 15, 2011 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: