953: For Group and 2 Commutative Elements of Different Orders, Order of Multiplication of 2 Elements Is Least Common Multiple of Orders

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that for group and 2 commutative elements of different orders, order of multiplication of 2 elements is least common multiple of orders

---

Topics

About: group

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof
• 3: Note

---

Starting Context

• The reader knows a definition of order of element of group.

---

Target Context

• The reader will have a description and a proof of the proposition that for any group and any 2 commutative elements of any different orders, the order of the multiplication of the 2 elements is the least common multiple of the orders.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$G$: $\in \{\text{ the groups }\}$
$g_1$: $\in G$, with $|(g_1)|=n_1$, where $n_1\in \mathbb{N}\setminus \{0\}$
$g_2$: $\in G$, with $|(g_2)|=n_2$, where $n_2\in \mathbb{N}\setminus \{0\}$
//

Statements:
(
$g_1{g}_2=g_2{g}_1$
$\wedge$
$n_1<n_2$
)
$⟹$
$|(g_1{g}_2)|=lcm(n_1,n_2)$
//

---

2: Proof

Whole Strategy: Step 1: let $l:=lcm(n_1,n_2)=n_1{m}_1=n_2{m}_2$ where $gcd(m_1,m_2)=1$; Step 2: see that $(g_1{g}_2{)}^l=1$ and see that $n:=|(g_1{g}_2)|$ is a divisor of $l$; Step 3: suppose that $n<l$, see that $n$ was a common divisor of $n_1$ and $n_2$, let $n_j=n{n}_j^{\prime }$, and find a contradiction from $((g_1{g}_2{)}^n{)}^{n_1^{\prime }}=1$.

Step 1:

Let $l:=lcm(n_1,n_2)=n_1{m}_1=n_2{m}_2$.

$gcd(m_1,m_2)=1$, because otherwise, $n_1{m}_1/gcd(m_1,m_2)=n_2{m}_2/gcd(m_1,m_2)$ would be a smaller common multiple.

Step 2:

$(g_1{g}_2{)}^l=g_1^l{g}_2^l$, because $g_1{g}_2=g_2{g}_1$, $=g_1^{n_1{m}_1}{g}_2^{n_2{m}_2}=(g_1^{n_1}{)}^{m_1}(g_2^{n_2}{)}^{m_2}=1^{m_1}{1}^{m_2}=11=1$.

So, $n:=|(g_1{g}_2)|$ is a divisor of $l$.

Step 3:

Let us suppose that $n<l$.

$n$ would be a common divisor of $n_1$ and $n_2$, because $gcd(m_1,m_2)=1$.

So, let $n_1=n{n}_1^{\prime }$ and $n_2=n{n}_2^{\prime }$.

$((g_1{g}_2{)}^n{)}^{n_1^{\prime }}=1^{n_1^{\prime }}=1$, but $((g_1{g}_2{)}^n{)}^{n_1^{\prime }}=(g_1{g}_2{)}^{n{n}_1^{\prime }}=(g_1{g}_2{)}^{n_1}=g_1^{n_1}{g}_2^{n_1}=1g_2^{n_1}=g_2^{n_1}$.

So, $g_2^{n_1}=1$, which is a contradiction against $|(g_2)|=n_2$, because $n_1<n_2$.

So, $n=l$.

---

3: Note

When $n_1=n_2$, $|(g_1{g}_2)|$ may not equal $lcm(n_1,n_2)$: as a counterexample, $g_1=g_2$ and $n_1=n_2=4$, then, $g_1{g}_2=g_1^2$, and $(g_1{g}_2{)}^2=(g_1^2{)}^2=g_1^4=1$, so, $|(g_1{g}_2)|=2\ne lcm(4,4)=4$.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>