Skip to main content Link Search Menu Expand Document (external link)

Quick review of examples of group actions

  • Dihedral group $D_{2n}$ acts on vertices of regular polygon with $n$ sides. The stabilizer of a vertex is of order two; cosets are in bijection with $n$ vertices. This group also acts on itself by conjugation. What are the orbits?

  • $\GL_{2}(\R)$ acts transitively on lines through the origin in $\R^{2}$. The stabilizer of the $x$-axis are the matrices of shape \(\left( \begin{matrix} a & b f\\ 0 & d\end{matrix} \right)\) The cosets are in bijection with points of the real projective line, with representatives \(\left[\begin{matrix} u & 1\end{matrix}\right]\) and \(\left[\begin{matrix} 1 & 0 \end{matrix}\right].\) The action on a coset representative is by linear fractional transformations in $u$.

  • The symmetric group acts on the set $\lbrace 1,\ldots, \rbrace n$ through its natural realization as bijections of this set to itself.

  • The symmetric group acts on its conjugacy classes. What can you say about $S_{4}$ acting on its conjugacy classes?

  • Let $F$ be a finite graph. The automorphisms of $F$ are the maps from $F$ to itself that preserve the edges (so connected vertices stay connected). Examples?

Proposition: Suppose that $G$ has order $n$ and $p$ is the smallest prime dividing $n$. Then any subgroup of $G$ of index $p$ is normal.

Proof: Consider the action of $G$ on the $p$ cosets of $H$. The kernel of this action is the set of $g\in G$ such that $gg_1H=g_1H$, which means that $gg_1=g_1h$ for some $h\in H$. This in turn means that $g\in g_1Hg_1^{-1}$. Since $g_1$ is arbitrary, we must have $g$ in the intersection of all conjugates $g_1Hg_1^{-1}$ as $g_1$ ranges over $G$. This is the largest normal subgroup of $G$ contained in $H$ – call it $K$.

The group $G/K$ acts faithfully on the cosets $G/H$, so it is isomorphic to a subgroup of $S_{p}$. Therefore $[G:K]$ divides $p!$. But $[G:K]=[G:H][H:K]=p[H:K]$ and if this is a divisor of $p!$ then $[H:K]$ can only have prime divisors less than $p$. Since $p$ is the smallest prime divisor of the order of $G$, this means $[H:K]=1$ so $K=H$ and $H$ is normal.

Remark: In the course of this we proved that, in general, the kernel of the action of $G$ on $G/H$ is the largest normal subgroup of $G$ contained in $H$.

Class equation and applications

Lemma: The stabilizer of an element $g\in G$ under conjugation is the centralizer of $\lbrace g\rbrace$.

Theorem: Let $G$ be a finite group. Let $G$ act on itself by conjugation, yielding a partition of $G$ into disjoint conjugacy classes $K_1,\ldots, K_g$. Choose a representative $g_{i}$ for each class. Then

\[\mid G\mid = \sum_{i=1}^{g}\mid K_i\mid = \sum_{i=1}^{g} [G:C_{G}(g_{i})].\]

Proposition: Let $G$ be a group of prime power order. Then $G$ has nontrivial center.

Corollary: If $G$ has order $p^{2}$ for some prime $p$, then $G$ is abelian.

Proof: If $xZ$ generates the quotient group, then every element of $G$ is of the form $x^{i}z^{j}$ with $z\in Z$. This forces $G$ to be abelian.

Conjugacy in $S_{n}$

The conjugacy classes in $S_{n}$ correspond to the cycle decompositions, and there is one class for each partition of $n$ as a sum of positive integers.

The centralizer of a cycle are the permutations which fix the integers appearing in the cycle.

The normalizer of a cycle was computed in the homework, at least in one case.