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 $G$.

Proof.

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

Since the identity element $e$ of $G$ satisfies $ea=a=ae$, it is in the centralizer $C_G(a)$.
Hence $C_G(a)$ is not an empty set. We show that $C_G(a)$ is closed under multiplications and inverses.

Let $g, h \in C_G(a)$. Then we have
\begin{align*}
(gh)a&=g(ha)\\
&=g(ah) && \text{since $h\in C_G(a)$}\\
&=(ga)h\\
&=(ag)h&& \text{since $g\in C_G(a)$}\\
&=a(gh).
\end{align*}
So $gh$ commutes with $a$ and thus $gh \in C_G(a)$.
Thus $C_G(a)$ is closed under multiplications.

Let $g\in C_G(a)$. This means that we have $ga=ag$.
Multiplying by $g^{-1}$ on the left and on the right, we obtain
\begin{align*}
g^{-1}(ga)g^{-1}=g^{-1}(ag)g^{-1},
\end{align*}
and thus we have
\[ag^{-1}=g^{-1}a.\] This implies that $g^{-1}\in C_G(a)$, hence $C_G(a)$ is closed under inverses.

Therefore, $C_G(a)$ is a subgroup of $G$.

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

We give two proofs for part (b).The first one is a more direct proof and the second one uses the orbit-stabilizer theorem.

The First Proof of (b).

By part (a), the centralizer $C_G(a)$ is a subgroup of the finite group $G$.
Hence the set of left cosets $G/C_G(a)$ is a finite set, and its order divides the order of $G$ by Lagrange’s theorem.

We prove that there is a bijective map from $G/C_G(a)$ to $\Cl(a)$.
Define the map $\phi:G/C_G(a) \to \Cl(a)$ by
\[\phi\left(\, gC_G(a) \,\right)=gag^{-1}.\]

We must show that it is well-defined.
For this, note that we have
\begin{align*}
gC_G(a)=hC_G(a) &\Leftrightarrow h^{-1}g\in C_G(a)\\
& \Leftrightarrow (h^{-1}g)a(h^{-1}g)^{-1}=a\\
& \Leftrightarrow gag^{-1}=hag^{-1}.
\end{align*}
This computation shows that the map $\phi$ is well-defined as well as $\phi$ is injective.
Since the both sets are finite sets, this implies that $\phi$ is bijective.
Thus, the order of the two sets is equal.

It yields that the order of $C_G(a)$ divides the order of the finite group $G$.

The Second Proof of (b). Use the Orbit-Stabilizer Theorem

We now move on to the alternative proof.
Consider the action of the group $G$ on itself by conjugation:
\[\psi:G\times G \to G, \quad (g,h)\mapsto g\cdot h=ghg^{-1}.\]

Then the orbit $\calO(a)$ of an element $a\in G$ under this action is
\[\calO(a)=\{ g\cdot a \mid g\in G\}=\{gag^{-1} \mid g\in G\}=\Cl(a).\]

Let $G_a$ be the stabilizer of $a$.
Then the orbit-stabilizer theorem for finite groups say that we have
\begin{align*}
|\Cl(a)|=|\calO(a)|=[G:G_a]=\frac{|G|}{|G_a|}
\end{align*}
and hence the order of $\Cl(a)$ divides the order of $G$.

Note that the stabilizer $G_a$ of $a$ is the centralizer $C_G(a)$ of $a$ since
\[G_a=\{g \in G \mid g\cdot a =a\}=\{g\in G \mid ga=ag\}=C_G(a).\] Click here if solved 40

The post The Order of a Conjugacy Class Divides the Order of the Group first appeared on Problems in Mathematics.

Let $P$ be a $p$-group acting on a finite set $X$.
Let
\[ X^P=\{ x \in X \mid g\cdot x=x \text{ for all } g\in P \}. \]

The prove that
\[|X^P|\equiv |X| \pmod{p}.\]

Proof.

Let $\calO(x)$ denote the orbit of $x\in X$ under the action of the group $P$.

Let $X^P=\{x_1, x_2, \dots, x_m\}$.
The orbits of an element in $X^p$ under the action of $P$ is the element itself, that is, $\calO(x_i)=\{x_i\}$ for $i=1,\dots, m$. Let $x_{m+1}, x_{m+2},\dots, x_n$ be representatives of other orbits of $X$.

Then we have the decomposition of the set $X$ into a disjoint union of orbits
\[X=\calO(x_1)\sqcup \cdots \sqcup \calO(x_m)\sqcup \calO(x_{m+1})\sqcup \cdots \sqcup \calO(x_n).\]

For $j=m+1, \dots, n$, the orbit-stabilizer theorem gives
\[|\calO(x_j)|=[P:\Stab_P(x_j)]=p^{\alpha_j}\] for some positive integer $\alpha_j$. Here $\alpha_j \neq 0$ otherwise $x_j \in X^P$.

Therefore we have
\begin{align*}
|X|&=\sum_{i=1}^m|\calO(x_i)|+\sum_{j=m+1}^n|\calO(x_j)|\\
&=\sum_{i=1}^m 1 +\sum_{j=m+1}^n p^{\alpha_j}\\
&=|X^P|+\sum_{j=m+1}^n p^{\alpha_j}\\
&\equiv |X^P| \pmod{p}.
\end{align*}
This completes the proof.

The post $p$-Group Acting on a Finite Set and the Number of Fixed Points first appeared on Problems in Mathematics.

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

Proof.

Let $\calO$ be the orbit of $e_n$ in $\F_p^n$.
We claim that $\calO=\F_p^n \setminus \{0\}$, hence
\[|\calO|=p^n-1.\]

To prove the claim, let $a_1 \in \F_p^n$ be a nonzero vector.
Then we can extend this vector to a basis of $\F_p^n$, that is, there is $a_2, \dots, a_n \in \F_p^n$ such that $a_1,\ a_2, \dots, a_n$ is a basis of $\F_P^n$.
Since they are a basis the matrix $A=[a_1 \dots a_n]$ is invertible, that is , $A \in G_n$.
We have
\[Ae_n=a_1\] Thus $a_1\in \calO$. It is clear that $0 \not \in \calO$. Thus we proved the claim.

Next we show that
\[|\Stab_{G_n}(e_n)|=|G_{n-1}|\cdot p^{n-1}. \tag{*} \] Note that $A \in \Stab_{G_n}(e_n)$ if and only if $A e_n=e_n$.
Thus $A$ is of the form
\[ \left[\begin{array}{r|r}
1 & A_2 \\ \hline
\mathbf{0} & A_1
\end{array} \right], \] where $A_1$ is an $(n-1)\times (n-1)$ matrix, $A_2$ is a $1\times (n-1)$ matrix , and $\mathbf{0}$ is the $(n-1) \times 1$ zero matrix.
Since $A$ is invertible, the matrix $A_1$ must be invertible as well, hence $A_1 \in G_{n-1}$.
The matrix $A_2$ can be anything.
Thus there are $|G_{n-1}|$ choices for $A_1$ and $p^{n-1}$ choices for $A_2$.
In total, there are $|G_{n-1}|p^{n-1}$ possible choices for $A \in \Stab_{G_n}(e_n)$. This proves (*).

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

When $n=1$, we have
\[|G_1|=|\F_p\setminus \{0\}|=p-1=p\left(1-\frac{1}{p} \right).\]

Now we assume that the formula is true for $n-1$.
By the orbit-stabilizer theorem, we have
\[ |G_n: \Stab_{G_n}(e_n)|=|\calO|.\] Since $G_n$ is finite, we have
\begin{align*}
|G_n|&=|\Stab_{G_n}(e_n)||\calO|\\
&=(p^n-1)|G_{n-1}|p^{n-1}\\
&=(p^n-1)p^{n-1}\cdot p^{(n-1)^2}\prod_{i=1}^{n-1} \left(1-\frac{1}{p^i} \right) \text{ by the induction hypothesis}\\
&=p^n\left(1-\frac{1}{p^n} \right)p^{n-1}p^{n^2-2n+1} \prod_{i=1}^{n-1} \left(1-\frac{1}{p^i} \right) \\
&=p^{n^2}\prod_{i=1}^{n} \left(1-\frac{1}{p^i} \right).
\end{align*}
Thus the formula is true for $n$ as well.

By induction, the formula is true for any $n$.

The post Group of Invertible Matrices Over a Finite Field and its Stabilizer first appeared on Problems in Mathematics.