# Abstract Nonsense

## 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 $G$ (not necessarily a $p$-group) and any $H\subseteq G$ we call the set

$\text{ }$

$\displaystyle \left\{g\in G:gHg^{-1}=H\right\}$

$\text{ }$

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

Theorem: Let $G$ be a group such that $|G|=p^n$ where $p$ is prime and $n\in\mathbb{N}$. Then:

$\text{ }$

\displaystyle \begin{aligned}&\mathbf{(1)}\quad\mathcal{Z}\left(G\right)\textit{ is non-trivial}\\ &\mathbf{(2)}\quad\mathcal{Z}\left(G\right)\cap N\textit{ is non-trivial for any non-trivial }N\unlhd G\\ &\mathbf{(3)}\quad \textit{If }H

$\text{ }$

Proof:

$\mathbf{(1)}$: Let $\Gamma$ be a set of one representative from each conjugacy class of $G$ with more than one element. By the class equation we have that

$\text{ }$

$\displaystyle p^n-\sum_{r\in\Gamma}\left(G:G_r\right)=\left|\mathcal{Z}\left(G\right)\right|$

$\text{ }$

note though that for each $r\in\Gamma$ we have that $\left(G:G_r\right)\left|G_r\right|=p^n$ so that $\left(G:G_r\right)\mid p^n$. Thus,

$\text{ }$

$\displaystyle p\mid \sum_{r\in\Gamma}\left(G:G_r\right)$

$\text{ }$

and thus consequently

$\text{ }$

$\displaystyle p\mid p^n-\sum_{r\in\Gamma}\left(G:G_r\right)=\left|\mathcal{Z}\left(G\right)\right|$

$\text{ }$

and since $\left|\mathcal{Z}\left(G\right)\right|\geqslant 1$ it follows that $\left|\mathcal{Z}\left(G\right)\right|=pn$ for some $n\in\mathbb{N}$. The conclusion follows.

$\text{ }$

$\mathbf{(2)}$: Recall from our proof the class equation that

$\text{ }$

$\displaystyle G=\mathcal{Z}\left(G\right)\;\uplus\;\biguplus_{r\in\Gamma}C(r)$

$\text{ }$

and thus

$\text{ }$

$\displaystyle N=\left(\mathcal{Z}\left(G\right)\cap N\right)\;\uplus\;\biguplus_{r\in\Gamma}\left(N\cap C(r)\right)$

$\text{ }$

and thus

$\text{ }$

$\displaystyle \left|N\right|=\left|\mathcal{Z}(G)\cap N\right|+\sum_{r\in\Gamma}\#\left(N\cap C(x)\right)$

$\text{ }$

notice though that if $x\in N$ then

$\text{ }$

$\displaystyle C(x)=\left\{gxg^{-1}:g\in G\right\}\subseteq gNg^{-1}=N$

$\text{ }$

(since $N\unlhd G$) and if $x\notin N$ then $C(x)\cap N=\varnothing$. It follows then that

$\text{ }$

$\displaystyle \left|N\right|=\left|\mathcal{Z}\left(G\right)\cap N\right|+\sum_{r\in\Gamma\cap N}\left(G:G_n\right)$

$\text{ }$

but using the same logic as before, noting that $|N|\mid p^n$, we may conclude that $p\mid\left|\mathcal{Z}\left(G\right)\right|$ and thus $\mathcal{Z}\left(G\right)$ is non-trivial.

$\text{ }$

$\mathbf{(3)}$: This last result follows by noticing that $\left(G:H\right)=p$ which is the smallest prime dividing $|G|$ and the conclusion follows from an earlier result.

$\blacksquare$

$\text{ }$

Corollary: Every group of order $p^2$ where $p$ is a prime is abelian.

Proof: By the above theorem we have that $\mathcal{Z}\left(G\right)$ is non-trivial and so by Lagrange’s Theorem we have that $\displaystyle \left|\mathcal{Z}\left(G\right)\right|=p,p^2$. Assume that $\left|\mathcal{Z}\left(G\right)\right|=p$ then $\left|G/\mathcal{Z}\left(G\right)\right|=p$ and thus (from previous theorem) we may conclude that $G/\mathcal{Z}\left(G\right)$ is cyclic. But, from an earlier theorem we may then conclude that $G$ is abelian and thus $\left|\mathcal{Z}\left(G\right)\right|=p^2$ which is a contradiction. Thus, it follows that $\left|\mathcal{Z}\left(G\right)\right|=p^2$ and thus $\mathcal{Z}\left(G\right)=G$ from where the conclusion follows. $\blacksquare$

$\text{ }$

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

January 6, 2011 -

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