IMC 2001 · Day 1 · P2

1. There exist integers $u$ and $v$ such that $us + vt = 1$. Since $ab = ba$, we obtain $$ab = (ab)^{us + vt} = (ab)^{us} \left( (ab)^t \right)^v = (ab)^{us} e = (ab)^{us} = a^{us} (b^s)^u = a^{us} e = a^{us}.$$ Therefore, $b^r = e b^r = a^r b^r = (ab)^r = a^{usr} = (a^r)^{us} = e$. Since $xr + ys = 1$ for suitable integers $x$ and $y$, $$b = b^{xr + ys} = (b^r)^x (b^s)^y = e.$$ It follows similarly that $a = e$ as well.

2. This is not true. Let $a = (123)$ and $b = (34567)$ be cycles of the permutation group $S_7$ of order 7. Then $ab = (1234567)$ and $a^3 = b^5 = (ab)^7 = e$.