Unique Composition of Group Element whose Order is Product of Coprime Integers

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Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $g \in G$ be an element of $g$.

Let:

$\order g = m n$

where:

$\order g$ denotes the order of $g$ in $G$
$m$ and $n$ are coprime integers.


Then $g$ can be expressed uniquely as the product of two commuting elements $a$ and $b$ of order $m$ and $n$ respectively.


Proof

Let $g_1 = g^n$ and $g_2 = g^m$.

By Powers of Group Element Commute, $g_1$ and $g_2$ commute.

We have:

${g_1}^m = g^{n m} = e$
${g_2}^n = g^{m n} = e$

It follows that:

the order of $g_1$ is $m$
the order of $g_2$ is $n$

otherwise if either $g_1$ or $g_2$ were of a smaller order then $g$ would also be of a smaller order.


By Bézout's Identity:

$\exists u, v \in \Z: u n + v m = 1$

as $m \perp n$.

Thus:

$g = g^{u n + v m} = \paren {g^n}^u \paren {g^m}^v = {g_1}^u {g_2}^v$

Also by Bézout's Identity:

$u \perp m$

and:

$v \perp n$

Thus by Order of Group Element equals Order of Coprime Power:

$\order { {g_1}^u} = m$

and:

$\order { {g_2}^v} = n$

We have that $g_1$ and $g_2$ commute.

So by Commutativity of Powers in Group, ${g_1}^u$ and ${g_2}^v$ also commute.

Putting $a = g_1^u$ and $b = g_2^v$, we have $a$ and $b$ which satisfy the required conditions.


It remains to prove uniqueness.

Suppose that:

$(1): \quad g = r_1 r_2 = s_1 s_2$

where:

$r_1$ and $r_2$ commute
$s_1$ and $s_2$ commute
$\order {r_1} = \order {s_1} = m$
$\order {r_2} = \order {s_2} = n$


Raising $(1)$ to the $n u$th power:

$g^{n u} = {r_1}^{n u} {r_2}^{n u} = {s_1}^{n u} {s_2}^{n u}$

and so:

\(\ds {r_1}^{n u} {r_2}^{n u}\) \(=\) \(\ds {s_1}^{n u} {s_2}^{n u}\)
\(\ds {r_1}^{n u} \paren { {r_2}^n}^u\) \(=\) \(\ds {s_1}^{n u} \paren { {s_2}^n}^u\)
\(\ds {r_1}^{n u} e^u\) \(=\) \(\ds {s_1}^{n u} e^u\) as $\order {r_2} = \order {s_2} = n$
\(\ds \leadsto \ \ \) \(\ds {r_1}^{n u}\) \(=\) \(\ds {s_1}^{n u}\)
\(\ds \leadsto \ \ \) \(\ds {r_1}^{1 - m v}\) \(=\) \(\ds {s_1}^{1 - m v}\)
\(\ds \leadsto \ \ \) \(\ds r_1 \paren { {r_1}^m}^{-v}\) \(=\) \(\ds s_1 \paren { {s_1}^m}^{-v}\)
\(\ds \leadsto \ \ \) \(\ds r_1 e^{-v}\) \(=\) \(\ds s_1 e^{-v}\)
\(\ds \leadsto \ \ \) \(\ds r_1\) \(=\) \(\ds s_1\)

It follows directly from $(1)$ that $r_2 = s_2$.

Hence the result.

$\blacksquare$


Also see


Sources

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