1990 Putnam B4

Let $G$ be a finite group of order $n$ generated by $a$ and
$b$. Prove or disprove: there is a sequence
$$g_1, g_2, g_3, \dots, g_{2n}$$
such that

(1) every element of $G$ occurs exactly twice, and

(2) $g_{i+1}$ equals $g_i a$ or $g_i b$ for $i = 1, 2, \dots,

2n$. (Interpret$g_{2n+1}$as$g_1$.)

令 $G$ 为 $a$ 生成的 $n$ 阶有限群，并且

$b$。证明或反驳：存在一个序列

$$

g_1, g_2, g_3, \dots, g_{2n}

$$

这样

(1) $G$ 的每个元素恰好出现两次，并且

(2) 当 $i = 1, 2, \dots 时，$g_{i+1}$等于$g_i a$或$g_i b$，
2n$。 (将$g_{2n+1}$解释为$g_1$。)