Let $G$ be a finite group and let $S$ be a non-empty set.
Suppose that $G$ acts on $S$ freely and transitively.
Prove that $|G|=|S|$. That is, the number of elements in $G$ and $S$ are the same.

Definition (Free and Transitive Group Action)

• A group action of a group $G$ on a set $S$ is called free if whenever we have
  \[gs=hs\] for some $g, h\in G$ and $s\in S$, this implies $g=h$.
• A group action of a group $G$ on a set $S$ is called transitive if for each pair $s, t\in S$ there exists an element $g\in G$ such that
  \[gs=t.\]

Proof.

We simply denote by $gs$ the action of $g\in G$ on $s\in S$.

Since $S$ is non-empty, we fix an element $s_0 \in S$. Define a map
\[\phi: G \to S\] by sending $g\in G$ to $gs_0 \in S$.
We prove that the map $\phi$ is bijective.

Suppose that we have $\phi(g)=\phi(h)$ for some $g, h\in G$.
Then it gives $gs_0=hs_0$, and since the action is free this implies that $g=h$.
Thus $\phi$ is injective.

To show that $\phi$ is surjective, let $s$ be an arbitrary element in $S$.
Since the action is transitive, there exists $g\in G$ such that $gs_0=s$.
Hence we have $\phi(g)=s$, and $\phi$ is surjective.

Therefore the map $\phi:G\to S$ is bijective, and we conclude that $|G|=|S|$.

The post If a Finite Group Acts on a Set Freely and Transitively, then the Numbers of Elements are the Same first appeared on Problems in Mathematics.

Prove that every finite group of order $72$ is not a simple group.

Definition.

A group $G$ is said to be simple if the only normal subgroups of $G$ are the trivial group $\{e\}$ or $G$ itself.

Hint.

Let $G$ be a group of order $72$.

Use the Sylow’s theorem and determine the number of Sylow $3$-subgroups of $G$.

If there is only one Sylow $3$-subgroup, then it is a normal subgroup, hence $G$ is not simple.

If there are more than one, consider the action of $G$ on those Sylow $3$-subgroups given by conjugation.
Then consider the induced permutation representation.

For a review of the Sylow’s theorem, check out the post “Sylow’s Theorem (summary)“.

Proof.

Observe the prime factorization $72=2^3\cdot 3^2$.
Let $G$ be a group of order $72$.

Let $n_3$ be the number of Sylow $3$-subgroups in $G$.
By Sylow’s theorem, we know that $n_3$ satisfies
\begin{align*}
&n_3\equiv 1 \pmod{3} \text{ and }\\
&n_3 \text{ divides } 8.
\end{align*}
The first condition gives $n_3$ could be $1, 4, 7, \dots$.
Only $n_3=1, 4$ satisfy the second condition.

Now if $n_3=1$, then there is a unique Sylow $3$-subgroup and it is a normal subgroup of order $9$.
Hence, in this case, the group $G$ is not simple.

It remains to consider the case when $n_3=4$.
So there are four Sylow $3$-subgroups of $G$.
Note that these subgroups are not normal by Sylow’s theorem.

The group $G$ acts on the set of these four Sylow $3$-subgroups by conjugation.
Hence it affords a permutation representation homomorphism
\[f:G\to S_4,\] where $S_4$ is the symmetric group of degree $4$.

By the first isomorphism theorem, we have
\begin{align*}
G/\ker f < S_4.
\end{align*}
Thus, the order of $G/\ker f$ divides the order of $S_4$.
Since $|S_4|=4!=2^3\cdot 3$, the order $|\ker f|$ must be divisible by $3$ (otherwise $|G/\ker f$|$ does not divide $|S_4|$), hence $\ker f$ is not the trivial group.

We claim that $\ker f \neq G$.
If $\ker f=G$, then it means that the action given by the conjugation by any element $g\in G$ is trivial.

That is, $gPg^{-1}=P$ for any $g\in G$ and for any Sylow $3$-subgroup $P$.
Since those Sylow $3$-subgroups are not normal, this is a contradiction.
Thus, $\ker f \neq G$.

Since a kernel of a homomorphism is a normal subgroup, this yields that $\ker f$ is a nontrivial proper normal subgroup of $G$, hence $G$ is not a simple group.

The post Every Group of Order 72 is Not a Simple Group first appeared on Problems in Mathematics.

Let $G$ be a finite group of order $p^n$, where $p$ is a prime number and $n$ is a positive integer.
Suppose that $H$ is a subgroup of $G$ with index $[G:P]=p$.
Then prove that $H$ is a normal subgroup of $G$.

(Michigan State University, Abstract Algebra Qualifying Exam)

Proof.

Let $G/H$ be the set of left cosets of $H$.
Then the group $G$ acts on $G/H$ by the left multiplication.
This action induces the permutation representation homomorphism
\[\phi: G\to S_{G/H},\] where $S_{G/H}$ is the symmetric group on $G/H$.
For each $g\in G$, the map $\phi(g):G/H \to G/H$ is given by $x\mapsto gx$.

By the first isomorphism theorem, we have
\begin{align*}
G/\ker(\phi) \cong \im(\phi) < S_{G/H}.
\end{align*}
This implies the order $|G/\ker(\phi)|$ divides the order $|S_{G/H}|=p!$.

Since
\[|G/\ker(\phi)|=\frac{|G|}{|\ker(\phi)|}=\frac{p^n}{|\ker(\phi)|}\] and $p!$ contains only one factor of $p$, we must have either $|\ker(\phi)|=p^n$ or $|\ker(\phi)|=p^{n-1}$.

Note that if $g\in \ker(\phi)$, then $\phi(g)=\id:G/H \to G/H$.
This yields that $gH=H$, and hence $g\in H$.
As a result, we have $\ker(\phi) \subset H$.

Since the index of $H$ is $p$, the order of $H$ is $p^{n-1}$.
Thus we conclude that $|\ker(\phi)|=p^{n-1}$ and
\[\ker(\phi)=H.\]

Since every kernel of a group homomorphism is a normal subgroup, the subgroup $H=\ker(\phi)$ is a normal subgroup of $G$.

The post A Subgroup of Index a Prime $p$ of a Group of Order $p^n$ is Normal first appeared on Problems in Mathematics.

Let $G$ be a finite group of order $231=3\cdot 7 \cdot 11$.
Prove that every Sylow $11$-subgroup of $G$ is contained in the center $Z(G)$.

Hint.

Prove that there is a unique Sylow $11$-subgroup of $G$, and consider the action of $G$ on the Sylow $11$-subgroup by conjugation.

Check out the post “Sylow’s Theorem (summary)” for a review of Sylow’s theorem.

Proof.

We first claim that there is a unique Sylow $11$-subgroup of $G$.
Let $n_{11}$ be the number of Sylow $11$-subgroups in $G$.

By Sylow’s theorem, we know that
\begin{align*}
&n_{11}\equiv 1 \pmod{11}\\
&n_{11}|21.
\end{align*}
By the first condition, $n_{11}=1, 12, 23 \cdots$ and only $n_{11}=1$ divides $21$.
Thus, we have $n_{11}=1$ and there is only one Sylow $11$-subgroup $P_{11}$ in $G$, and hence it is normal in $G$.

Now we consider the action of $G$ on the normal subgroup $P_{11}$ given by conjugation.
The action induces the permutation representation homomorphism
\[\psi:G\to \Aut(P_{11}),\] where $\Aut(P_{11})$ is the automorphism group of $P_{11}$.

Note that $P_{11}$ is a group of order $11$, hence it is isomorphic to the cyclic group $\Zmod{11}$.
Recall that
\[\Aut(\Zmod{11})\cong (\Zmod{11})^{\times}\cong \Zmod{10}.\]

The first isomorphism theorem gives
\begin{align*}
G/\ker(\psi) \cong \im(\psi) < \Aut(P_{11})\cong \Zmod{10}.
\end{align*}

Hence the order of $G/\ker(\psi)$ must be a divisor of $10$.
Since $|G|=231=3\cdot 7 \cdot 11$, the only possible way for this is $|G/\ker(\psi)|=1$ and thus $\ker(\psi)=G$.

This implies that for any $g\in G$, the automorphism $\psi(g): P_{11}\to P_{11}$ given by $h\mapsto ghg^{-1}$ is the identity map.
Thus, we have $ghg^{-1}=h$ for all $g\in G$ and $h\in H$.
It yields that $P_{11}$ is in the center $Z(G)$ of $G$.

The post Every Sylow 11-Subgroup of a Group of Order 231 is Contained in the Center $Z(G)$ first appeared on Problems in Mathematics.

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 $G$ be a group and let $H$ be a subgroup of finite index. Then show that there exists a normal subgroup $N$ of $G$ such that $N$ is of finite index in $G$ and $N\subset H$.

Proof.

The group $G$ acts on the set of left cosets $G/H$ by left multiplication.
Hence it induces the permutation representation $\rho: G \to S_n$, where $n=|G:H|$.
(Note that a permutation representation is a group homomorphism.)
Let $N=\ker \rho$ be the kernel of the homomorphism $\rho$. Then $N \triangleleft G$.

By the first isomorphism theorem, the quotient group $G/N$ is isomorphic to a subgroup of $S_n$. In particular, $G/N$ is a finite group, hence the index $[G:N]$ is finite.

Finally, we show that $N \subset H$.
For any $x \in N=\ker \rho$, we have $x(gH)=gH$ for any $g \in G$.
In particular we have $xH=H$, hence $x \in H$.

The post Subgroup of Finite Index Contains a Normal Subgroup of Finite Index 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.

Let $G$ be a finite group of order $n$ and suppose that $p$ is the smallest prime number dividing $n$.

Then prove that any subgroup of index $p$ is a normal subgroup of $G$.
 

Hint.

Consider the action of the group $G$ on the left cosets $G/H$ by left multiplication.

Proof.

Let $H$ be a subgroup of index $p$.
Then the group $G$ acts on the left cosets $G/H$ by left multiplication.

It induces the permutation representation $\rho: G \to S_p$.

Let $K=\ker \rho$ be the kernel of $\rho$.
Since $kH=H$ for $k\in K$, we have $K\subset H$.
Let $[H:K]=m$.

By the first isomorphism theorem, the quotient group $G/K$ is isomorphic to the subgroup of $S_p$, thus $[G:K]$ divides $|S_p|=p!$ by Lagrange’s theorem.
Since $[G:K]=[G:H][H:K]=pm$, we have $pm|p!$ and hence $m|(p-1)!$.

If $m$ has a prime factor $q$, then $q\geq p$ since the minimality of $p$ but the factors of $(p-1)!$ are only prime numbers less than $p$.
Thus $m|(p-1)!$ implies that $m=1$, hence $H=K$. Therefore $H$ is normal since a kernel is always normal.

The post A Subgroup of the Smallest Prime Divisor Index of a Group is Normal first appeared on Problems in Mathematics.