Let $G$ be an abelian group and let $H$ be the subset of $G$ consisting of all elements of $G$ of finite order. That is,
\[H=\{ a\in G \mid \text{the order of $a$ is finite}\}.\]

Prove that $H$ is a subgroup of $G$.

Proof.

Note that the identity element $e$ of $G$ has order $1$, hence $e\in H$ and $H$ is not an empty set.

To show that $H$ is a subgroup of $G$, we need to show that $H$ is closed under multiplications and inverses.

Let $a, b\in H$.
By definition of $H$, the orders of $a, b$ are finite.
So let $m, n \in \N$ be the orders of $a, b$, respectively:
We have
\[a^m=e \text{ and } b^n=e.\]

Then we have
\begin{align*}
(ab)^{mn}&=a^{mn}b^{mn} && \text{since $G$ is abelian}\\
&=(a^m)^n(b^n)^m\\
&=e^ne^m=e.
\end{align*}

This implies that the order of $ab$ is at most $mn$, hence the order of $ab$ is finite.
Thus $ab\in H$ for any $a, b\in H$.

Next, consider any $a\in H$. We want to show that the inverse $a^{-1}$ also lies in $H$.
Let $m \in \N$ be the order of $a$.
Then we have
\begin{align*}
(a^{-1})^m=(a^m)^{-1}=e^{-1}=e.
\end{align*}
This implies that the order of $a^{-1}$ is also finite, and hence $a^{-1}\in H$.

Therefore we have proved that $H$ is closed under multiplications and inverses.
Hence $H$ is a subgroup of $G$.

The post Elements of Finite Order of an Abelian Group form a Subgroup first appeared on Problems in Mathematics.

Let $G$ be an abelian group.
Let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively.
Prove that there exists an element $c$ in $G$ such that the order of $c$ is the least common multiple of $m$ and $n$.

Also determine whether the statement is true if $G$ is a non-abelian group.

Hint.

First, consider the case when $m$ and $n$ are relatively prime.

Proof.

When $m$ and $n$ are relatively prime

Recall that if the orders $m, n$ of elements $a, b$ of an abelian group are relatively prime, then the order of the product $ab$ is $mn$.
(For a proof, see the post “Order of the Product of Two Elements in an Abelian Group“.)

So if $m, n$ are relatively prime, then we can take $c=ab\in G$ and $c$ has order $mn$, which is the least common multiple.

The general Case

Now we consider the general case.

Let $p_i$ be the prime factors of either $m$ or $n$.
Then write prime factorizations of $m$ and $n$ as
\[m=\prod_{i}p_i^{\alpha_i} \text{ and } n=\prod_{i} p_i^{\beta_i}.\] Here $\alpha_i$ and $\beta_i$ are nonzero integers (could be zero).
Define
\[m’=\prod_{i: \alpha_i \geq \beta_i}p_i^{\alpha_i} \text{ and } n’=\prod_{i: \beta_i> \alpha_i} p_i^{\beta_i}.\]

(For example, if $m=2^3\cdot 3^2\cdot 5$ and $n=3^2\cdot 7$ then $m’=2^3\cdot 3^2\cdot 5$ and $n’=7$.)

Note that $m’\mid m$ and $n’\mid n$, and also $m’$ and $n’$ are relatively prime. The least common multiple $l$ of $m$ and $n$ is given by
\[l=m’n’\]

Consider the element $a’:=a^{m/m’}$. We claim that the order of $a’$ is $m’$.

Let $k$ be the order of the element $a’$. Then we have
\begin{align*}
e=(a’)^k=(a^{\frac{m}{m’}})^k=a^{mk/m’},
\end{align*}
where $e$ is the identity element in the group $G$.
This yields that $m$ divides $mk/m’$ since $m$ is the order of $a$.
It follows that $m’$ divides $k$.

On the other hand, we have
\begin{align*}
(a^{m/m’})^{m’}=a^m=e,
\end{align*}
and hence $k$ divides $m’$ since $k$ is the order of the element $a^{m/m’}$.

As a result, we have $k=m’$.
So the order of $a’$ is $m’$.

Similarly, the order of $b’:=b^{n/n’}$ is $n’$.

The orders of elements $a’$ and $b$ are $m’$ and $n’$, and they are relatively prime.
Hence we can apply the first case and we conclude that the element $a’b’$ has order
\[m’n’=l.\] Thus, we can take $c=a’b’$.

The Case When $G$ is a Non-Abelian Group

Next, we show that if $G$ is a non-abelian group then the statement does not hold.

For example, consider the symmetric group $S_3$ with three letters.
Let
\[a=(1\,2\,3) \text{ and } b=(1 \,2).\]

Then the order of $a$ is $3$ and the order of $b$ is $2$.
The least common multiple of $2$ and $3$ is $6$.
However, the symmetric group $S_3$ have no elements of order $6$.

Hence the statement of the problem does not hold for non-abelian groups.

Related Question.

For a solution of this problem, see the post “Order of Product of Two Elements in a Group“.

The post The Existence of an Element in an Abelian Group of Order the Least Common Multiple of Two Elements first appeared on Problems in Mathematics.

Let $G$ be a group. Let $a$ and $b$ be elements of $G$.
If the order of $a, b$ are $m, n$ respectively, then is it true that the order of the product $ab$ divides $mn$? If so give a proof. If not, give a counterexample.

Proof.

We claim that it is not true. As a counterexample, consider $G=S_3$, the symmetric group of three letters.
Let $a=(1\, 2), b=(1 \,3)$ be transposition elements in $S_3$.
The orders of $a$ and $b$ are both $2$.

Consider the product
\[ab=(1\, 2)(1 \,3)=(1 \, 3 \, 2).\] Then it is straightforward to check that the order of $ab$ is $3$, which does not divide $4$ (the product of orders of $a$ and $b$).

Therefore, the group $G=S_3$ and elements $a=(1\, 2), b=(1 \,3)\in G$ serve as a counterexample.

Remark. (Abelian group case)

If we further assume that $G$ is an abelian group, then the statement is true.
Here is the proof if $G$ is abelian.

Let $e$ be the identity element of $G$.
\begin{align*}
(ab)^{mn} &=a^{mn}b^{mn} && \text{ since $G$ is abelian}\\
&=(a^m)^n(b^n)^m\\
&=e^n e^m && \text{since the order of $a, b$ are $m, n$ respectively}\\
&=e.
\end{align*}
Thus the order of $ab$ divides $mn$.

Related Question.

If the group is abelian, then the statement is true.

See the post “Order of the Product of Two Elements in an Abelian Group” for a proof of this problem.

More generally, we can prove the following.

A proof of this problem is given in the post “The Existence of an Element in an Abelian Group of Order the Least Common Multiple of Two Elements“.

The post Order of Product of Two Elements in a Group first appeared on Problems in Mathematics.

Prove that if $G$ is a finite group of even order, then the number of elements of $G$ of order $2$ is odd.

Proof.

First observe that for $g\in G$,
\[g^2=e \iff g=g^{-1},\] where $e$ is the identity element of $G$.
Thus, the identity element $e$ and the elements of order $2$ are the only elements of $G$ that are equal to their own inverse elements.

Hence, each element $x$ of order greater than $2$ comes in pairs $\{x, x^{-1}\}$.
So we have
\begin{align*}
&G=\\
&\{e\}\cup \{\text{ elements of order $2$ } \} \cup \{x_1, x_1^{-1}, x_2, x_2^{-1}, \dots, x_k, x_k^{-1}\},
\end{align*}
where $x_i$ are elements of order greater than $2$ for $i=1,2, \dots, k$.

As we noted above, the elements $x_i, x_i^{-1}$ are distinct.
Thus the third set contains an even number of elements.

Therefore we have
\begin{align*}
&\underbrace{G}_{\text{even}}=\\
&\underbrace{\{e\}}_{\text{odd}}\cup \{\text{ elements of order $2$ } \}\cup \underbrace{\{x_1, x_1^{-1}, x_2, x_2^{-1}, \dots, x_k, x_k^{-1}\}}_\text{even}
\end{align*}
It follows that the number of elements of $G$ of order $2$ must be odd.

If the Order of a Group is Even, then it has a Non-Identity Element of Order 2

The consequence of the problem yields that the number of elements of order $2$ is odd, in particular, it is not zero.

Hence we obtain:

The post If the Order of a Group is Even, then the Number of Elements of Order 2 is Odd first appeared on Problems in Mathematics.

Let $A$ be an abelian group and let $T(A)$ denote the set of elements of $A$ that have finite order.

(a) Prove that $T(A)$ is a subgroup of $A$.

(The subgroup $T(A)$ is called the torsion subgroup of the abelian group $A$ and elements of $T(A)$ are called torsion elements.)

(b) Prove that the quotient group $G=A/T(A)$ is a torsion-free abelian group. That is, the only element of $G$ that has finite order is the identity element.

Proof.

(a) $T(A)$ is a subgroup of $A$

We write the group operation multiplicatively.
Let $x, y\in T(A)$. Then $x, y$ have finite order, hence there exists positive integers $m, n$ such that $x^m=e, y^n=e$, where $e$ is the identity element of $A$. Then we have
\begin{align*}
(xy)^{mn}&=x^{mn}y^{mn} \qquad \text{ (since $A$ is abelian)}\\
&=(x^m)^n(y^m)^n=e^me^n=e.
\end{align*}
Therefore the element $xy$ has also finite order, hence $xy \in T(A)$.

Also, we have
\begin{align*}
(x^{-1})^m=(x^m)^{-1}=e^{-1}=e.
\end{align*}
Hence the inverse $x^{-1}$ of $x$ has finite order, hence $x^{-1}\in T(A)$.

Therefore, the subset $T(A)$ is closed under group operation and inverse, hence $T(A)$ is a subgroup of $A$.

(b) $A/T(A)$ is a torsion-free abelian group

Since $A$ is an abelian group, the quotient $G=A/T(A)$ is also an abelian group.
For $a\in A$, let $\bar{a}=aT(A)$ be an element of $G=A/T(A)$. Suppose that $\bar{a}$ has finite order in $G$. We want to prove that $\bar{a}=\bar{e}$ the identity element of $G$.

Since $\bar{a}$ has finite order, there exists a positive integer $n$ such that
\[\bar{a}^n=\bar{e}.\] This implies that
\[a^nT(A)=T(A)\] and thus $a^n\in T(A)$.

Since each element of $T(A)$ has finite order by definition, there exists a positive integer $m$ such that $(a^n)^m=e$.
It follows from $a^{nm}=e$ that $a$ has finite order, and thus $a\in T(A)$.
Therefore we have
\[\bar{a}=aT(A)=T(A)=\bar{e}.\]

We have proved that any element of $G=A/T(A)$ that has finite order is the identity, hence $G$ is the torsion-free abelian subgroup of $G$.

The post Torsion Subgroup of an Abelian Group, Quotient is a Torsion-Free Abelian Group first appeared on Problems in Mathematics.

Let $G$ be a group. Suppose that the order of nonidentity element of $G$ is $2$.
Then show that $G$ is an abelian group.

Proof.

Let $x$ and $y$ be elements of $G$. Then we have
\[1=(xy)^2=(xy)(xy).\]

Multiplying the equality by $yx$ from the right, we obtain
\begin{align*}
yx&=(xy)(xy)(yx)\\
&=xyxy^2x\\
&=xyx^2 \quad (\text{ since } y^2=1)\\
&=xy \quad (\text{ since } x^2=1).
\end{align*}
Thus we obtain $xy=yx$ for any elements $x, y \in G$. Thus the group $G$ is an abelian group.

The post If Every Nonidentity Element of a Group has Order 2, then it's an Abelian Group first appeared on Problems in Mathematics.

Let $G$ be an abelian group with the identity element $1$. Let $a, b$ be elements of $G$ with order $m$ and $n$, respectively.
If $m$ and $n$ are relatively prime, then show that the order of the element $ab$ is $mn$.

Proof.

Let $r$ be the order of the element $ab$.
Since we have
\begin{align*}
(ab)^{mn}&=a^{mn}b^{mn} \quad \text{ (since } G \text{ is an abelian group)}\\
&=(a^m)^n(b^n)^m\\
&=1
\end{align*}

since $a^m=1$ and $b^n=1$.
This implies that the order $r$ of $ab$ divides $mn$, that is, we have
\[ r |mn. \tag{*}\]

Now, since $r$ is the order of $ab$ we have
\[1=(ab)^r=a^rb^r.\] Then we have
\begin{align*}
1=1^n=a^{rn}b^{rn}=a^{rn}
\end{align*}

since $b^n=1$. This yields that the order $m$ of the element $a$ divides $rn$.

Since $m$ and $n$ are relatively prime, this implies that we have
\[m|r.\]

Similarly (switch the role of $n$ and $m$), we obtain
\[n|r.\] Thus we have
\[mn|r \tag{**}\] since $m$ and $n$ are relatively prime.

From (*) and (**), we have $r=mn$, and hence the order of the element $ab$ is $mn$.

Related Question.

As a generalization of this problem, try the following problem.

A proof of this problem is given in the post “The Existence of an Element in an Abelian Group of Order the Least Common Multiple of Two Elements“.

Also See the post “Order of product of two elements in a group” for a similar problem about the order of elements in a non-abelian group.

The post Order of the Product of Two Elements in an Abelian Group first appeared on Problems in Mathematics.