Equal Powers of Finite Order Element

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Theorem

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

Let $g \in G$ be of finite order.

Let $\order g = k$.


Then:

$g^r = g^s \iff k \divides \paren {r - s}$


Proof

Necessary Condition

Suppose that $k \divides \paren {r - s}$.

From the definition of divisor:

$k \divides \left({r - s}\right) \implies \exists t \in \Z: r - s = k t$

So:

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

Thus:

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

$\Box$


Sufficient Condition

Let $g^r = g^s$.

Then:

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


By the Division Theorem:

$r - s = q k + t$

for some $q \in \Z, 0 \le t < k$.

Thus:

$e = g^{r - s} = g^{k q + t} = \paren {g^k}^q g^t = e^q g^t = g^t$


So by the definition of $k$:

$\paren {t < k} \land \paren {e = g^t} \implies t = 0$

So:

$r - s = q k + 0 = q k \implies k \divides \paren {r - s}$

$\blacksquare$


Sources