$𝑎 𝑏$ and $𝑏 𝑎$ have the same order

In any group, $| 𝑎 𝑏 | = | 𝑏 𝑎 |$ , so if $(𝑎 𝑏) 𝑛 = 𝑒$ then also $(𝑏 𝑎) 𝑛 = 𝑒$ . #m/thm/group

Proof

If $(𝑎 𝑏) 𝑛 = 𝑒$ for some $𝑛 > 0$ then

𝑎 (𝑏 𝑎) 𝑛 - 1 𝑏 = 𝑒 (𝑏 𝑎) 𝑛 - 1 = 𝑎 - 1 𝑏 - 1 (𝑏 𝑎) 𝑛 = 𝑒 ◻

#state/tidy | #lang/en | #SemBr

𝑎𝑏 and 𝑏𝑎 have the same order