Equal Powers of Finite Order Element
Jump to navigation
Jump to search
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:
\(\ds g^r\) | \(=\) | \(\ds g^{s + k t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds g^s g^{k t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds g^s \paren {g^k}^t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds g^s \paren e^t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds g^s\) |
$\Box$
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
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 5.4$. Cyclic groups
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $3$: Elementary consequences of the definitions: Proposition $3.10$