04

題目

If $G$ is a group in which $(a\cdot b)^i = a^{i} \cdot b^{i}$, for three consecutive integers $i$ for all $a, b$ in $G$, show that $G$ is abelian.

解答

Since $(ab)^{i+2} = a\cdot (ba)^{i+1}\cdot b$, and

$$a^{i+2}\cdot b^{i+2} = a\cdot a^{i+1}\cdot b^{i+1}\cdot b = a\cdot (ab)^{i+1}\cdot b$$

then $(ab)^{i+1} = (ba)^{i+1}$.

In the same way, $(ab)^i = (ba)^i$. Then

$$ab\cdot(ba)^i = ab\cdot(ab)^i = (ab)^{i+1} = (ba)^{i+1} = ba\cdot(ba)^i$$

hence $a\cdot b = b\cdot a.$

note

$a, b, (ba)^{i}$ has inverse, because they are in $G$.