# Abstract Nonsense

## 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 $H\unlhd G$.

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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

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

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and

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$e\cdot kH=(ek)H=kH$

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

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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

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

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where $i_g$ is the inner automorphism induced by $g$.

Proof: By definition we have that

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

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Also,

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\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}

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$\blacksquare$

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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

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

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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 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. $\blacksquare$

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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

January 4, 2011 -

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