Problem 628

Let $G$ be a group of order $57$. Assume that $G$ is not a cyclic group.

Then determine the number of elements in $G$ of order $3$.
 

Proof.

Observe the prime factorization $57=3\cdot 19$.
Let $n_{19}$ be the number of Sylow $19$-subgroups of $G$.

By Sylow’s theorem, we know that
\[n_{19} \equiv 1 \pmod{19} \text{ and } n_{19} \mid 3.\] It follows that $n_{19}=1$.

Now, observe that if $g\in G$, then the order of $g$ is $1$, $3$, or $19$. Note that since $G$ is not a cyclic group, the order of $g$ cannot be $57$.
As there is exactly one Sylow $19$-subgroup $P$, any element that is not in $P$ must have order $3$.

Therefore, the number of elements of order $3$ is $57-19=38$.

Remark.

Note that there are $18$ elements of order $19$ and the identity element is the only element of order $1$.

The post Determine the Number of Elements of Order 3 in a Non-Cyclic Group of Order 57 first appeared on Problems in Mathematics.

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.

Is it possible that each element of an infinite group has a finite order?
If so, give an example. Otherwise, prove the non-existence of such a group.

Solution.

We give an example of a group of infinite order each of whose elements has a finite order.
Consider the group of rational numbers $\Q$ and its subgroup $\Z$.
The quotient group $\Q/\Z$ will serve as an example as we verify below.

Note that each element of $\Q/\Z$ is of the form
\[\frac{m}{n}+\Z,\] where $m$ and $n$ are integers.

This implies that the representatives of $\Q/\Z$ are rational numbers in the interval $[0, 1)$.
There are infinitely many rational numbers in $[0, 1)$, and hence the order of the group $\Q/\Z$ is infinite.

On the other hand, as each element of $\Q/\Z$ is of the form $\frac{m}{n}+\Z$ for $m, n\in \Z$, we have
\[n\cdot \left(\, \frac{m}{n}+\Z \,\right)=m+\Z=0+\Z\] because $m\in \Z$.
Thus the order of the element $\frac{m}{n}+\Z$ is at most $n$.
Hence the order of each element of $\Q/\Z$ is finite.

Therefore, $\Q/\Z$ is an infinite group whose elements have finite orders.

The post Example of an Infinite Group Whose Elements Have Finite Orders 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 $G$ a finite group and let $H$ and $K$ be two distinct Sylow $p$-group, where $p$ is a prime number dividing the order $|G|$ of $G$.

Prove that the product $HK$ can never be a subgroup of the group $G$.

Hint.

Use the following fact.

Proof.

Let $p^{\alpha}$ is the highest power of $p$ that divides $|G|$.
That is, we have
\[|G|=p^{\alpha}n,\] where $p$ does not divide the integer $n$.

Then the orders of the Sylow $p$-subgroups $H, K$ are $p^{\alpha}$.

Since the intersection $H\cap K$ is a subgroup of $H$, the order of $H \cap K$ is $p^{\beta}$ for some integer $\beta \leq \alpha$ by Lagrange’s theorem.
As $H$ and $K$ are distinct subgroups, we must have $\beta < \alpha$.

Then the number of elements of the product $HK$ is
\begin{align*}
|HK|&=\frac{|H| |K|}{|H \cap K|}\\[6pt] &=\frac{p^{\alpha} p^{\alpha}}{p^{\beta}}=p^{2\alpha-\beta}.
\end{align*}
Since $\beta < \alpha$, we have $2\alpha-\beta > \alpha$.

It follows that the product $HK$ cannot be a subgroup of $G$ since otherwise the order $|HK|=p^{2\alpha-\beta}$ divides $|G|$ by Lagrange’s theorem but $p^{\alpha}$ is the highest power of $p$ that divides $G$.

The post The Product of Distinct Sylow $p$-Subgroups Can Never be a Subgroup first appeared on Problems in Mathematics.

Let $G$ be a finite group and let $A, B$ be subsets of $G$ satisfying
\[|A|+|B| > |G|.\] Here $|X|$ denotes the cardinality (the number of elements) of the set $X$.
Then prove that $G=AB$, where
\[AB=\{ab \mid a\in A, b\in B\}.\]

Proof.

Since $A, B$ are subsets of the group $G$, we have $AB\subset G$.
Thus, it remains to show that $G\subset AB$, that is any element $g\in G$ is of the form $ab$ for some $a\in A$ and $b\in B$.
This is equivalent to finding $a\in A$ and $b\in B$ such that $gb^{-1}=a$.

Consider the subset
\[B^{-1}:=\{b^{-1} \mid b \in B\}.\] Since taking the inverse gives the bijective map $B \to B^{-1}$, $b \mapsto b^{-1}$, we have $|B|=|B^{-1}|$.

Also consider the subset
\[gB^{-1}=\{gb^{-1} \mid b\in B\}.\] Note that multiplying by $g$ and by its inverse $g^{-1}$ give the bijective maps
\[B^{-1} \to gB^{-1}, b^{-1} \mapsto gb^{-1} \text{ and } gB^{-1} \to B^{-1}, gb^{-1} \mapsto b^{-1}.\] Hence we have
\[ |B|=|B^{-1}|=|gB^{-1}|.\]

Since $A$ and $gB^{-1}$ are both subsets in $G$ and we have by assumption that
\[|A|+|gB^{-1}|=|A|+|B| > |G|,\] the intersection $A\cap gB^{-1}$ cannot be empty.

Therefore, there exists $a \in A\cap gB^{-1}$, and thus $a\in A$ and $a=gb^{-1}$ for some $b\in B$.
As a result we obtain $g=ab$.
It yields that $G\subset AB$, and we have $G=AB$ as a consequence.

Related Question.

As an application, or use the similar technique, try the following

See the post↴
Each Element in a Finite Field is the Sum of Two Squares
for a proof of this problem.

The post If Two Subsets $A, B$ of a Finite Group $G$ are Large Enough, then $G=AB$ first appeared on Problems in Mathematics.

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.

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

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