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 group. Suppose that the number of elements in $G$ of order $5$ is $28$.

Determine the number of distinct subgroups of $G$ of order $5$.

Solution.

Let $g$ be an element in $G$ of order $5$.
Then the subgroup $\langle g \rangle$ generated by $g$ is a cyclic group of order $5$.
That is, $\langle g \rangle=\{e, g, g^2, g^3, g^4\}$, where $e$ is the identity element in $G$.

Note that the order of each non-identity element in $\langle g \rangle$ is $5$.

Also, if $h$ is another element in $G$ of order $5$, then we have either $\langle g \rangle=\langle h \rangle$ or $\langle g \rangle \cap \langle h \rangle = \{e\}$.
This follows from the fact that the intersection $\langle g \rangle \cap \langle h \rangle$ is a subgroup of the order $5$ group $\langle g \rangle$, and thus the order of $\langle g \rangle \cap \langle h \rangle$ is either $5$ or $1$.

On the other hand, if $H$ is a subgroup of $G$ of order $5$, then every non-identity element in $H$ has order $5$.

These observations imply that each subgroup of order $5$ contains exactly $4$ elements of order $5$ and each element of order $5$ appears in exactly one of such subgroups.

As there are $28$ elements of order $5$, there are $28/4=7$ subgroups of order $5$.

The post If There are 28 Elements of Order 5, How Many Subgroups of Order 5? first appeared on Problems in Mathematics.

Let $G$ be a group and let $H_1, H_2$ be subgroups of $G$ such that $H_1 \not \subset H_2$ and $H_2 \not \subset H_1$.

(a) Prove that the union $H_1 \cup H_2$ is never a subgroup in $G$.

(b) Prove that a group cannot be written as the union of two proper subgroups.

Proof.

Prove that the union $H_1 \cup H_2$ is never a subgroup in $G$.

Seeking a contradiction, let us assume that the union $H_1 \cup H_2$ is a subgroup of $G$.
Since $H_1 \not \subset H_2$, there exists an element $a\in H_1$ such that $a\notin H_2$.
Similarly, as $H_2 \not \subset H_1$, there exists an element $b\in H_2$ such that $b\notin H_1$.

As we are assuming $H_1 \cup H_2$ is a group, we have $ab\in H_1 \cup H_2$.
It follows that either $ab \in H_1$ or $ab \in H_2$.

If $ab \in H_1$, then we have
\[b=a^{-1}(ab) \in H_1\] as both $a^{-1}$ and $ab$ are elements in the subgroup $H_1$.
This contradicts our choice of the element $b$.

Similarly, if $ab \in H_2$, we have
\[ a=(ab)b^{-1} \in H_2,\] which contradicts the choice of $a$.

In either case, we reached a contradiction.
Thus, we conclude that the union $H_1 \cup H_2$ is not a subgroup of $G$.

(b) Prove that a group cannot be written as the union of two proper subgroups.

This is a special case of part (a).

If a group $G$ is a union of two proper subgroup $H_1$ and $H_2$, then we must have $H_1 \not \subset H_2$ and $H_2 \not \subset H_1$, otherwise $G=H_1$ or $G=H_2$ and this is impossible as $H_1, H_2$ are proper subgroups.
Then $G=H_1\cup H_2$ is a subgroup of $G$, which is prohibited by part (a).

Thus, any group cannot be a union of proper subgroups.

The post Union of Two Subgroups is Not a Group 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.

Prove that every cyclic group is abelian.

Proof.

Let $G$ be a cyclic group with a generator $g\in G$.
Namely, we have $G=\langle g \rangle$ (every element in $G$ is some power of $g$.)

Let $a$ and $b$ be arbitrary elements in $G$.
Then there exists $n, m\in \Z$ such that $a=g^n$ and $b=g^m$.

It follows that
\begin{align*}
ab&=g^ng^m=g^{n+m}=g^mg^n=ba.
\end{align*}

Hence we obtain $ab=ba$ for arbitrary $a, b\in G$.
Thus $G$ is an abelian group.

The post Every Cyclic Group is Abelian first appeared on Problems in Mathematics.

Suppose that $p$ is a prime number greater than $3$.
Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$.

(a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$.

(b) Determine the index $[G : S]$.

(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$.

Proof.

(a) Prove that $S=\{x^2\mid x\in G\}$ is a subgroup of $G$.

Consider the map $\phi:G \to G$ defined by $\phi(x)=x^2$ for $x\in G$.
Then $\phi$ is a group homomorphism. In fact, for any $x, y \in G$, we have
\begin{align*}
\phi(xy)=(xy)^2=x^2y^2=\phi(x)\phi(y)
\end{align*}
as $G$ is an abelian group.

By definition of $\phi$, the image is $\im(\phi)=S$.
Since the image of a group homomorphism is a group, we conclude that $S$ is a subgroup of $G$.

(b) Determine the index $[G : S]$.

By the first isomorphism theorem, we have
\[G/\ker(\phi)\cong S.\]

If $x\in \ker(\phi)$, then $x^2=1$.
It follows that $(x-1)(x+1)=0$ in $\Zmod{p}$.
Since $\Zmod{p}$ is an integral domain, it follows that $x=\pm 1$ and $\ker(\phi)=\{\pm 1\}$.

Thus, $|S|=|G/\ker(\phi)|=(p-1)/2$ and hence the index is
\[[G:S]=|G|/|S|=2.\]

(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$.

Since $-1\notin S$ and $[G:S]=2$, we have the decomposition
\[G=S\sqcup (-S).\] Suppose that an element $a$ in $G$ is not in $S$.

Then, we have $a\in -S$.
Thus, there exists $b\in S$ such that $a=-b$.
It follows that $-a=b\in S$. Therefore, we have either $a\in S$ or $-a\in S$.

The post The Set of Square Elements in the Multiplicative Group $(\Zmod{p})^*$ first appeared on Problems in Mathematics.

Let $G$ be a finite group. Let $S$ be the set of elements $g$ such that $g^5=e$, where $e$ is the identity element in the group $G$.

Prove that the number of elements in $S$ is odd.

Proof.

Let $g\neq e$ be an element in the group $G$ such that $g^5=e$.
As $5$ is a prime number, this yields that the order of $g$ is $5$.

Consider the subgroup $\langle g \rangle$ generated by $g$.
As the order of $g$ is $5$, the order of the subgroup $\langle g \rangle$ is $5$.

If $h\neq e$ is another element in $G$ such that $h^5=e$, then we have either $\langle g \rangle=\langle h \rangle$ or $\langle g \rangle \cap \langle h \rangle=\{e\}$ as the intersection of these two subgroups is a subgroup of $\langle g \rangle$.

It follows that $S$ is the union of subgroups of order $5$ that intersect only at the identity element $e$.
Thus the number of elements in $S$ are $4n+1$ for some nonnegative integer $n$.

The post The Number of Elements Satisfying $g^5=e$ in a Finite Group is Odd first appeared on Problems in Mathematics.

Let $m$ and $n$ be positive integers such that $m \mid n$.

(a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined.

(b) Prove that $\phi$ is a group homomorphism.

(c) Prove that $\phi$ is surjective.

(d) Determine the group structure of the kernel of $\phi$.

Proof.

(a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined.

To show that $\phi$ is well-defined, we need show that the value of $\phi$ does not depends on the choice of representative $a$.
So suppose that $a+n\Z=a’+n\Z$ so that $a$ and $a’$ are two representatives for the same element.
This yields that $a-a’$ is divisible by $n$.

Now, $a+n\Z$ is mapped to $a+m\Z$ by $\phi$. On the other hand, $a’+n\Z$ is mapped to $a+m\Z$ by $\phi$.
Since $a-a’$ is divisible by $n$ and $m \mid n$, it follows that $a-a’$ is divisible by $m$.
This implies that $a+m\Z=a’+m\Z$.
This prove that $\phi$ does not depend on the choice of the representative, and hence $\phi$ is well-defined.

(b) Prove that $\phi$ is a group homomorphism.

Let $a+n\Z$, $b+n\Z$ be two elements in $\Zmod{n}$. Then we have
\begin{align*}
&\phi\left(\, (a+n\Z)+(b+n\Z) \,\right)\\
&=\phi\left(\, (a+b)+n\Z) \,\right) &&\text{by addition in $\Zmod{n}$}\\
&=(a+b)+m\Z &&\text{by definition of $\phi$}\\
&=(a+m\Z)+(b+m\Z)&&\text{by addition in $\Zmod{m}$}\\
&=\phi(a+n\Z)+\phi(b+n\Z) &&\text{by definition of $\phi$}.
\end{align*}

Hence $\phi$ is a group homomorphism.

(c) Prove that $\phi$ is surjective.

For any $c+m\Z \in \Zmod{m}$, we pick $c+n\Z\in \Zmod{n}$.
Then as $\phi(c+n\Z)=c+m\Z$, we see that $\phi$ is surjective.

(d) Determine the group structure of the kernel of $\phi$.

If $a+n\Z\in \ker(\phi)$, then we have $0+m\Z=\phi(a+n\Z)=a+m\Z$.
This implies that $m\mid a$.
On the other hand, if $m\mid a$, then $\phi(a+n\Z)=a+m\Z=0+m\Z$ and $a+n\Z\in \ker(\phi)$.

It follows that
\[\ker(\phi)=\{mk+n\Z \mid k=0, 1, \dots, l-1\},\] where $l$ is an integer such that $n=ml$.

Thus, $\ker(\phi)$ is a group of order $l$.
Since $\ker(\phi)$ is a subgroup of the cyclic group $\Zmod{n}$, we know that $\ker(\phi)$ is also cyclic.
Thus
\[\ker(\phi)\cong \Zmod{l}.\]

Another approach

Here is a more direct proof of this result.
Define a map $\psi:\Z\to \ker(\phi)$ by sending $k\in \Z$ to $mk+n\Z$.
It is straightforward to verify that $\psi$ is a surjective group homomorphism and the kernel of $\psi$ is $\ker(\psi)=l\Z$.
It follows from the first isomorphism theorem that
\[\Zmod{l}= \Z/\ker(\psi) \cong \im(\psi)=\ker(\phi). \] Click here if solved 124

The post Group Homomorphism from $\Z/n\Z$ to $\Z/m\Z$ When $m$ Divides $n$ 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.