Tagged: group action

The Order of a Conjugacy Class Divides the Order of the Group

Problem 455

Let $G$ be a finite group.
The centralizer of an element $a$ of $G$ is defined to be
\[C_G(a)=\{g\in G \mid ga=ag\}.\]

A conjugacy class is a set of the form
\[\Cl(a)=\{bab^{-1} \mid b\in G\}\] for some $a\in G$.

(a) Prove that the centralizer of an element of $a$ in $G$ is a subgroup of the group $G$.

(b) Prove that the order (the number of elements) of every conjugacy class in $G$ divides the order of the group $G$.

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Group of Invertible Matrices Over a Finite Field and its Stabilizer

Problem 108

Let $\F_p$ be the finite field of $p$ elements, where $p$ is a prime number.
Let $G_n=\GL_n(\F_p)$ be the group of $n\times n$ invertible matrices with entries in the field $\F_p$. As usual in linear algebra, we may regard the elements of $G_n$ as linear transformations on $\F_p^n$, the $n$-dimensional vector space over $\F_p$. Therefore, $G_n$ acts on $\F_p^n$.

Let $e_n \in \F_p^n$ be the vector $(1,0, \dots,0)$.
(The so-called first standard basis vector in $\F_p^n$.)

Find the size of the $G_n$-orbit of $e_n$, and show that $\Stab_{G_n}(e_n)$ has order $|G_{n-1}|\cdot p^{n-1}$.

Conclude by induction that
\[|G_n|=p^{n^2}\prod_{i=1}^{n} \left(1-\frac{1}{p^i} \right).\]

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