Equal Powers of Finite Order Element

Theorem

Let $G$ be a group whose identity is $e$.

Let $g \in G$ be of finite order, and let $\left|{g}\right| = k$.

Then:

$g^r = g^s \iff k \backslash \left({r - s}\right)$.

Proof

• First, suppose that $k \backslash \left({r - s}\right)$.

From the definition of divisor, $k \backslash \left({r - s}\right) \implies \exists t \in \Z: r - s = k t$.

So $g^{r - s} = g^{k t}$.

Thus:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle g^r\)	\(=\)	\(\displaystyle g^{s + k t}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle g^s g^{k t}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle g^s \left({g^k}\right)^t\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle g^s \left({e}\right)^t\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle g^s\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

• Now let $g^r = g^s$.

Then:

$g^{r-s} = g^r g^{-s} = g^s g^{-s} = e$

Now by the Division Theorem, we can say $r - s = q k + t$ for some $q \in \Z, 0 \le t < k$.

Thus:

$e = g^{r - s} = g^{k q + t} = \left({g^k}\right)^q g^t = e^q g^t = g^t$

So $\left({t < k}\right) \land \left({e = g^t}\right) \implies t = 0$ by the definition of $k$.

So:

$r - s = q k + 0 = q k \implies k \backslash \left({r - s}\right)$

$\blacksquare$

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 5.4$
• John F. Humphreys: A Course in Group Theory (1996): $\S 3$: Proposition $3.10$