Prove that every group of order $24$ has a normal subgroup of order $4$ or $8$.

Proof.

Let $G$ be a group of order $24$.
Note that $24=2^3\cdot 3$.
Let $P$ be a Sylow $2$-subgroup of $G$. Then $|P|=8$.

Consider the action of the group $G$ on the left cosets $G/P$ by left multiplication.
This induces a permutation representation homomorphism
\[\phi: G\to S_{G/P},\] where $S_{G/P}$ is a group of bijective maps (permutations) on $G/P$.

This homomorphism is defined by
\[\phi(g)(aP)=gaP\] for $g\in G$ and $aP\in G/P$.

Then by the first isomorphism theorem, we see that
\[G/\ker(\phi) \cong \im(\phi) < S_{G/P}.\] This implies that the order of $G/\ker(\phi)$ divides the order of $S_{G/P}$. Note that as $|G/P|=3$, we have $|S_{G/P}|=|S_3|=6$. Thus, we must have $4\mid |\ker{\phi}|$.

Also note that $\ker(\phi) < P$. To see this let $x\in \ker(\phi)$. Then we have \[xP=\phi(x)(P)=\id(P)=P.\] Here $\id$ is the identity map from $G/P$ to itself. Hence $x\in P$. It follows that $|\ker(\phi)|$ divides $|P|=8$.

Combining these restrictions, we see that $|\ker(\phi)|=4, 8$.
Being the kernel of a homomorphism, $\ker(\phi)$ is a normal subgroup of $G$.
Hence the group $G$ of order $24$ has a normal subgroup of order $4$ or $8$.

The post Every Group of Order 24 Has a Normal Subgroup of Order 4 or 8 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 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 $G$ be a simple group and let $X$ be a finite set.
Suppose $G$ acts nontrivially on $X$. That is, there exist $g\in G$ and $x \in X$ such that $g\cdot x \neq x$.
Then show that $G$ is a finite group and the order of $G$ divides $|X|!$.

Proof.

Since $G$ acts on $X$, it induces a permutation representation
\[\rho: G \to S_{X}.\]

Let $N=\ker(\rho)$ be the kernel of $\rho$.
Since a kernel is normal in $G$ and $G$ is simple, we have either $N=\{e\}$ or $N=G$.

If $N=G$, then for any $g\in G$ we have $\rho(g)$ is a trivial action, that is, $g\cdot x=x$ for any $X$.
This contradicts the assumption that $G$ acts nontrivially on $X$.
Hence we have $N=\{e\}$, and it follows that the homomorphism $\rho$ is injective.

Thus we have
\[G \cong \mathrm{im} (\rho) < S_{X}.\] Since $S_{X}$ is a finite group and $G$ is isomorphic to its subgroup, the group $G$ is finite.
By Lagrange’s theorem, the order $|G|=|\mathrm{im}(\rho)|$ of $G$ divides the order $|S_{X}|=|X|!$ of $S_{X}$.

The post Nontrivial Action of a Simple Group on a Finite Set first appeared on Problems in Mathematics.