# Alg7.17

Let $G$ be a group such that $(ab)^{i}=a^{i}b^{i}$ for all $a,b\in G$ and for three consecutive integers $i$. Prove $G$ is abelian.

We are given three equations. For all $a,b\in G$

(1) $(ab)^{i}=a^{i}b^{i}\,$

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

(3) $(ab)^{{i+2}}=a^{{i+2}}b^{{i+2}}\,$

From (1), we have $[(ab)^{i}]^{{-1}}=b^{{-i}}a^{{-i}}\,$.

From (2), by multiplying by the inverse found from (1), we have

(4) $ab=a^{{i+1}}b^{{i+1}}b^{{-i}}a^{{-i}}=a^{{i+1}}ba^{{-i}}\,$

From (3), again multiplying by the inverse found from (1), we have

(5) $abab=a^{{i+2}}b^{{i+2}}b^{{-i}}a^{{-i}}=a^{{i+2}}b^{2}a^{{-i}}\,$

Take (4) and mutliply by $b$ on the left and (5) and multiply by $a^{{-1}}$ on the left and we get $ba^{{i+1}}ba^{{-i}}=bab=a^{{i+1}}b^{2}a^{{-i}}\,$

Now cancelling the $ba^{{-i}}$ on the right, we get $ba^{{i+1}}=a^{{i+1}}b\,$

Therefore, for every $a\in G$, it is true that $a^{{i+1}}$ commutes with every element of $G$. Using that information and (4), we get $ab=a^{{i+1}}ba^{{-i}}=ba^{{i+1}}a^{{-i}}=ba\,$

Therefore, for every element of $G$ commutes with every other element of $G$.