Powers Up to Order In Group Are Distinct

Theorem

Given an element \(g\) in a group \(G\) where \(g\) has order \(n\), the elements \(\{g^0, g^1, g^2, \dots, g^{n - 1}\}\) are all distinct.

If \(g\) has infinite order, then all distinct powers are distinct.

Proof

Let \(g \in G\) be an element of order \(n\). Then, consider \(g^i\) and \(g^j\) for some \(i, j \in \{0, 1, 2, \dots, n - 1\}\), such that \(g^i = g^j\).

Then, we have that \(g^{i - j} = \mathrm{id}\).

However this implies that \(i - j\) must be a multiple of \(n\), but \(i, j \in \{0, 1, 2, \dots, n - 1\}\), which leaves only the possibility that \(i - j = 0\).

Therefore we have that \(g^i = g^j \implies i = j\) and therefore by the contrapositive \(i \neq j \implies g^i \neq g^j\).

In the infinite case, we know that if \(g^i = g^j\) then similarly \(i - j = 0\), since no non-zero power of \(g\) is the identity.