The Order of Elements in Group

Proposition 1

Let $G$ be a group, let $a\in G$, and suppose that the order of $a$ is $m$. Prove that:

(1) $a^n=e$ if and only if $m\mid n$.

(2) $a^k=a^h$ if and only if $k\equiv h \pmod m$.

Proof : (1) Sufficiency:

Assume that $n=qm$. Then

$$
a^n=a^{qm}=(a^m)^q=e^q=e.
$$

Necessity:

By the division algorithm, write

$$
n=mq+r,\quad 0\le r<m.
$$

If $r\ne 0$, then

$$
a^r=a^{n-mq}=a^n(a^m)^{-q}=e,
$$

but $r<m$, which contradicts the fact that $m$ is the order of $a$. Therefore $r=0$, and hence $m\mid n$.

(2)

$$
a^k=a^h
\iff a^{k-h}=e
\iff m\mid (k-h)
\iff k\equiv h \pmod m.
$$

Proposition 2

For any elements $a,b\in G$, denote by $o(a)$ the order of $a$ in $G$. Prove that:

(1) $o(a)=o(a^{-1})$.

(2) $o(a)=o(bab^{-1})$.

(3) $o(ab)=o(ba)$.

Proof : (1) Suppose that the order of $a$ is $m$, and the order of $a^{-1}$ is $n$.

On the one hand,

$$
(a^{-1})^m=(a^m)^{-1}=e,
$$

so $n\mid m$.

On the other hand,

$$
a^n=\bigl((a^{-1})^n\bigr)^{-1}=e,
$$

so $m\mid n$.

Therefore $m=n$.

(2) Suppose that the order of $a$ is $m$, and the order of $bab^{-1}$ is $n$.

On the one hand,

$$
(bab^{-1})^m=ba^mb^{-1}=beb^{-1}=e,
$$

so $n\mid m$.

On the other hand,

$$
a^n=b^{-1}(bab^{-1})^n b=b^{-1}eb=e,
$$

so $m\mid n$.

Therefore $m=n$.

(3) Suppose that the order of $ab$ is $m$, and the order of $ba$ is $n$.

On the one hand,

$$
(ba)^m=b(ab)^{m-1}a=b(ab)^{-1}(ab)^m a=b(ab)^{-1}ea=e,
$$

so $n\mid m$.

On the other hand, similarly we obtain $m\mid n$.

Therefore $m=n$.

Proposition 3

Let $G$ be a group, let $a\in G$, and suppose that the order of $a$ is $m$. Prove that:

(1) The order of $a^k$ is $\dfrac{m}{\gcd(m,k)}$.

(2) The order of $a^k$ is equal to the order of $a$ if and only if $\gcd(m,k)=1$.

Proof : (1) Suppose that the order of $a^k$ is $q$, and let $d=\gcd(m,k)$.

Then there exist $u,v\in \mathbb{Z}$ such that

$$
um+vk=d.
$$

Hence

$$
u\frac{m}{d}+v\frac{k}{d}=1,
$$

and therefore

$$
u\frac{mq}{d}+v\frac{kq}{d}=q.
$$

Since the order of $a^k$ is $q$, we have

$$
(a^k)^q=a^{kq}=e.
$$

Thus $m\mid kq$, and so

$$
\frac{m}{d}\mid \frac{kq}{d}.
$$

Clearly,

$$
\frac{m}{d}\mid \frac{umq}{d}.
$$

Therefore,

$$
\frac{m}{d}\mid \left(\frac{umq}{d}+\frac{vkq}{d}\right)=q.
$$

On the other hand,

$$
(a^k)^{\frac{m}{d}}=a^{k\frac{m}{d}}=(a^m)^{\frac{k}{d}}=e,
$$

so

$$
q\mid \frac{m}{d}.
$$

Hence

$$
q=\frac{m}{d}=\frac{m}{\gcd(m,k)}.
$$

(2)

$$
\text{the order of } a^k \text{ equals the order of } a
\iff \frac{m}{\gcd(m,k)}=m
\iff \gcd(m,k)=1.
$$

Proposition 4

Let $G$ be a group, and let $a,b\in G$ satisfy $ab=ba$. Suppose that the order of $a$ is $m$, the order of $b$ is $n$, and $\gcd(m,n)=1$. Prove that the order of $ab$ is $mn$.

Proof : Suppose that the order of $ab$ is $q$.

On the one hand,

$$
(ab)^{mn}=a^{mn}b^{mn}=e,
$$

so $q\mid mn$.

On the other hand,

$$
a^{nq}=a^{nq}e=a^{nq}b^{nq}=(ab)^{nq}=e,
$$

so $m\mid nq$. Since $\gcd(m,n)=1$, it follows that $m\mid q$.

Similarly, we have $n\mid q$. Since $\gcd(m,n)=1$, we obtain $mn\mid q$.

Therefore $q=mn$.

Proposition 5

In a group $G$, if $ab=ba$, the order of $a$ is $m$, and the order of $b$ is $n$, must the order of $ab$ be $[m,n]$? Here $[m,n]$ denotes the least common multiple of $m$ and $n$.

Not necessarily. A counterexample is as follows.

In $\mathbb{Z}_8$, we have $o(\bar 2)=4$ and $o(\bar 4)=2$, and

$$
\bar 2\cdot \bar 2=\bar 4,
$$

but

$$
[4,4]=4\ne 2.
$$

Proposition 6

In a group $G$, suppose that $ab=ba$, the order of $a$ is $m$, and the order of $b$ is $n$. Let $(a)$ and $(b)$ denote the cyclic subgroups generated by $a$ and $b$, respectively. If $(a)\cap (b)=\{e\}$, prove that the order of $ab$ is $[m,n]$, where $[m,n]$ denotes the least common multiple of $m$ and $n$.

Proof : Suppose that the order of $ab$ is $q$. On the one hand,

$$
(ab)^q=a^q b^q=e,
$$

so

$$
a^q=b^{-q}\in (a)\cap (b)=\{e\}.
$$

Hence

$$
a^q=b^q=e,
$$

and therefore

$$
m\mid q,\quad n\mid q.
$$

Thus

$$
[m,n]\mid q.
$$

On the other hand,

$$
(ab)^{[m,n]}=a^{[m,n]}b^{[m,n]}=e,
$$

so

$$
q\mid [m,n].
$$

Therefore

$$
q=[m,n].
$$

Proposition 7

In a group $G$, suppose that $ab=ba$, the order of $a$ is $m$, and the order of $b$ is $n$. Prove that there must exist an element in $G$ whose order is $[m,n]$, where $[m,n]$ denotes the least common multiple of $m$ and $n$.

Proof : Write the prime factorizations of $m$ and $n$ as follows:

$$
m=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_r^{\alpha_r}.
$$

$$
n=p_1^{\beta_1}p_2^{\beta_2}\cdots p_r^{\beta_r},
$$

where $\alpha_i,\beta_i\ge 0$. Without loss of generality, assume that

$$
\alpha_i\ge \beta_i,\quad 1\le i\le s;\qquad \alpha_i<\beta_i,\quad s+1\le i\le r.
$$

Then

$$
[m,n]=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_s^{\alpha_s}p_{s+1}^{\beta_{s+1}}\cdots p_r^{\beta_r}.
$$

Let

$$
c=a^{p_{s+1}^{\alpha_{s+1}}\cdots p_r^{\alpha_r}},
$$

$$
d=b^{p_1^{\beta_1}p_2^{\beta_2}\cdots p_s^{\beta_s}}.
$$

By the result of Proposition 3, the order of $c$ is

$$
p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_s^{\alpha_s},
$$

and the order of $d$ is

$$
p_{s+1}^{\beta_{s+1}}\cdots p_r^{\beta_r}.
$$

Thus the orders of $c$ and $d$ are relatively prime.

By the result of Proposition 4, the order of $cd$ is

$$
p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_s^{\alpha_s}p_{s+1}^{\beta_{s+1}}\cdots p_r^{\beta_r}=[m,n].
$$

The cover image of this article was taken in Irkutsk Oblast, Russia.