Let $G$ be a finite group and let $N$ be a normal subgroup of $G$.
Suppose that the order $n$ of $N$ is relatively prime to the index $|G:N|=m$.

(a) Prove that $N=\{a\in G \mid a^n=e\}$.

(b) Prove that $N=\{b^m \mid b\in G\}$.

Proof.

Note that as $n$ and $m$ are relatively prime integers, there exits $s, t\in \Z$ such that
\[sn+tm=1. \tag{*}\] Also, note that as the order of the group $G/N$ is $|G/N|=|G:N|=m$, we have
\[g^mN=(gN)^m=N\] for any $g \in G$ by Lagrange’ theorem, and thus
\[g^m\in N \tag{**}.\]

(a) Prove that $N=\{a\in G \mid a^n=e\}$.

Suppose $a\in \{a\in G \mid a^n=e\}$. Then we have $a^n=e$.
It follows that
\begin{align*}
a \stackrel{(*)}{=} a^{sn+tm}= a^{sn}a^{tm}=a^{tm}=(a^t)^m\in N
\end{align*}
by (**).
This proves that $\{a\in G \mid a^n=e\} \subset N$.

On the other hand, if $a\in N$, then we have $a^n=e$ as $n$ is the order of the group $N$.
Hence $N\subset \{a\in G \mid a^n=e\}$.

Putting together these inclusions yields that $N=\{a\in G \mid a^n=e\}$ as required.

(b) Prove that $N=\{b^m \mid b\in G\}$.

Let $b^m \in \{b^m \mid b\in G\}$. Then by (**), we know that $b^m\in N$.
Thus, we have $\{b^m \mid b\in G\}\subset N$.

On the other hand, let $a\in N$. Then we have $a^n=e$ as $n=|N|$.
Hence it follows that
\begin{align*}
a \stackrel{(*)}{=} a^{sn+tm}= a^{sn}a^{tm}=a^{tm}=b^m,
\end{align*}
where we put $b:=a^t$.
This implies that $a\in \{b^m \mid b\in G\}$, and hence we have $N \subset \{b^m \mid b\in G\}$.

So we see that $N=\{b^m \mid b\in G\}$ by these two inclusions.

The post Normal Subgroup Whose Order is Relatively Prime to Its Index first appeared on Problems in Mathematics.

Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup.
Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.

Then prove that $H$ is an abelian normal subgroup of odd order.

Proof.

The index of the subgroup $H$ in $G$ is $2$, hence $H$ is a normal subgroup.
(See the post “Any Subgroup of Index 2 in a Finite Group is Normal“.)

Also, the order of $H$ must be odd, otherwise $H$ contains an element of order $2$.
So it remains to prove that $H$ is abelian.

Let $a\in S$ be an element of order $2$.
As $a\notin H$, the left coset $aH$ is different from $H$.
Since the index of $H$ is $2$, we have $aH=G\setminus H=S$.
So for any $h\in H$, the order of $ah$ is $2$.

It follows that we have for any $h\in H$
\[e=(ah)^2=ahah,\] where $e$ is the identity element in $G$.

Equivalently, we have
\[aha^{-1}=h^{-1} \tag{*}\] for any $h\in H$.
(Remark that $a=a^{-1}$ as the order of $a$ is $2$.)

Using this relation, for any $h, k \in H$, we have
\begin{align*}
(hk)^{-1}&\stackrel{(*)}{=} a(hk)a^{-1}\\
&=(aha^{-1})(aka^{-1})\\
&\stackrel{(*)}{=}h^{-1}k^{-1}=(kh)^{-1}.
\end{align*}

As a result, we obtain $hk=kh$ for any $h, k$.
Hence the subgroup $H$ is abelian.

The post If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order first appeared on Problems in Mathematics.

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 group of order $12$. Prove that $G$ has a normal subgroup of order $3$ or $4$.

Hint.

Use Sylow’s theorem.
(See Sylow’s Theorem (Summary) for a review of Sylow’s theorem.)

Recall that if there is a unique Sylow $p$-subgroup in a group $GH$, then it is a normal subgroup in $G$.

Proof.

Since $12=2^2\cdot 3$, a Sylow $2$-subgroup of $G$ has order $4$ and a Sylow $3$-subgroup of $G$ has order $3$.
Let $n_p$ be the number of Sylow $p$-subgroups in $G$, where $p=2, 3$.
Recall that if $n_p=1$, then the unique Sylow $p$-subgroup is normal in $G$.

By Sylow’s theorem, we know that $n_2\mid 3$, hence $n_p=1, 3$.
Also by Sylow’s theorem, $n_3 \equiv 1 \pmod{3}$ and $n_3\mid 4$.
It follows that $n_3=1, 4$.

If $n_3=1$, then the unique Sylow $3$-subgroup is a normal subgroup of order $3$.

Suppose that $n_3=4$. Then there are four Sylow $3$-subgroup in $G$.
The order of each Sylow $3$-subgroup is $3$, and the intersection of two distinct Sylow $3$-subgroups intersect trivially (the intersection consists of the identity element) since every nonidentity element has order $3$.
Hence two elements of order $3$ in each Sylow $3$-subgroup are not included in other Sylow $3$-subgroup.

Thus, there are totally $4\cdot 2=8$ elements of order $3$ in $G$.
Since $|G|=12$, there are $12-8=4$ elements of order not equal to $3$.

Since any Sylow $2$-subgroup contains four elements, these elements fill up the remaining elements.
So there is just one Sylow $2$-subgroup, and hence it is a normal subgroup of order $4$.

In either case, the group $G$ has a normal subgroup of order $3$ or $4$.

The post Every Group of Order 12 Has a Normal Subgroup of Order 3 or 4 first appeared on Problems in Mathematics.

Let $N$ be a normal subgroup of a group $G$.
Suppose that $G/N$ is an infinite cyclic group.

Then prove that for each positive integer $n$, there exists a normal subgroup $H$ of $G$ of index $n$.

Hint.

Use the fourth (or Lattice) isomorphism theorem.

Proof.

Let $n$ be a positive integer.
Since $G/N$ is a cyclic group, let $g$ be a generator of $G/N$.
So we have $G/N=\langle g\rangle$.
Then $\langle g^n \rangle$ is a subgroup of $G/N$ of index $n$.

By the fourth isomorphism theorem, every subgroup of $G/N$ is of the form $H/N$ for some subgroup $H$ of $G$ containing $N$.
Thus we have $\langle g^n \rangle=H/N$ for some subgroup $H$ in $G$ containing $N$.

Since $G/N$ is cyclic, it is in particular abelian.
Thus $H/N$ is a normal subgroup of $G/N$.

The fourth isomorphism theorem also implies that $H$ is a normal subgroup of $G$, and we have
\begin{align*}
[G:H]=[G/N : H/N]=n.
\end{align*}
Hence $H$ is a normal subgroup of $G$ of index $n$.

The post If the Quotient is an Infinite Cyclic Group, then Exists a Normal Subgroup of Index $n$ 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 $H$ be a subgroup of a group $G$.
Suppose that for each element $x\in G$, we have $x^2\in H$.

Then prove that $H$ is a normal subgroup of $G$.

(Purdue University, Abstract Algebra Qualifying Exam)

Proof.

To show that $H$ is a normal subgroup of $G$, we prove that
\[ghg^{-1}\in H\] for any $g\in G$ and $h\in H$.

For any $g\in G$ and $h\in H$ we have
\begin{align*}
&ghg^{-1}\\
&=g^2g^{-1}hg^{-1} &&\text{since $g=g^2g^{-1}$}\\
&=g^2g^{-1}hg^{-1}hh^{-1} &&\text{since $e=hh^{-1}$}\\
&=g^2(g^{-1}h)^2h^{-1}. \tag{*}
\end{align*}

It follows from the assumption that the elements $g^2$ and $(g^{-1}h)^2$ are in $H$.
Since $h\in H$, the inverse $h^{-1}$ is also in $H$.
Thus the expression in (*) is the product of elements in $H$, hence it is in $H$.

Thus, we have proved that $ghg^{-1}\in H$ for all $g\in G$, $h\in H$.
Therefore, the subgroup $H$ is a normal subgroup in $G$.

The post If Squares of Elements in a Group Lie in a Subgroup, then It is a Normal Subgroup 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.

Prove that every group of order $20449$ is an abelian group.

Outline of the Proof

Note that $20449=11^2 \cdot 13^2$.
Let $G$ be a group of order $20449$.
We prove by Sylow’s theorem that there are a unique Sylow $11$-subgroup and a unique Sylow $13$-subgroup of $G$.
Hence $G$ is the direct product of these Sylow subgroups.

Since these Sylow subgroups are of order $11^2$ and $13^2$, respectively, they are abelian.
Since the direct product of abelian groups is abelian, the group $G$ is abelian.

Proof.

Observe that $20449=11^2 \cdot 13^2$.
Let $G$ be a group of order $20449$.

Let $n_{11}$ be the number of Sylow $11$-subgroups of $G$.
By Sylow’s theorem, $n_{11}$ satisfies
\begin{align*}
&n_{11}\equiv 1 \pmod{11} \text{ and }\\
&n_{11} \text{ divides } 13^2.
\end{align*}

The second condition yields that $n_{11}$ could be $1, 13, 13^2$.
Among these numbers, only $n_{11}=1$ satisfies the first condition.
So there is a unique Sylow $11$-subgroup $P_{11}$ of $G$, hence $P_{11}$ is a normal subgroup of $G$.

Similarly, let $n_{13}$ be the number of Sylow $13$-subgroups of $G$.
Sylow’s theorem yields that $n_{13}$ satisfies:
\begin{align*}
&n_{13}\equiv 1 \pmod{13} \text{ and }\\
&n_{13} \text{ divides } 11^2.
\end{align*}
From the second condition, we see that $n_{13}$ could be $1, 11, 13$.
Among these numbers, only $n_{13}=1$ satisfies the first condition.
So there is a unique Sylow $13$-subgroup $P_{13}$ of $G$, hence $P_{13}$ is a normal subgroup of $G$.

Note that the orders of $P_{11}$ and $P_{13}$ are $11^2$ and $13^2$, respectively.
The intersection of $P_{11}$ and $P_{13}$ is the trivial group.
Thus, we have
\begin{align*}
|G|=\frac{|P_{11}P_{13}|}{|P_{11}\cap P_{13}|}=|P_{11}P_{13}|.
\end{align*}
This yields that $G=P_{11}P_{13}$.

In summary, we have

• Sylow subgroups $P_{11}$ and $P_{13}$ are normal in $G$.
• $P_{11}\cap P_{13}=\{e\}$.
• $G=P_{11}P_{13}$.

These implies that $G$ is the direct product of $P_{11}$ and $P_{13}$:
\[G=P_{11}\times P_{13}.\]

Recall that every group of order $p^2$ for some prime number $p$ is an abelian group.
Thus, $P_{11}$ and $P_{13}$ are both abelian group.
Since the direct product of abelian groups is abelian, we conclude that the group $G=P_{11}\times P_{13}$ is abelian.

The post Every Group of Order 20449 is an Abelian Group first appeared on Problems in Mathematics.