Alg7.17

From Exampleproblems

Let $G$ be a group such that $(a b) i = a i b i$ 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) $(ab)^i=a^ib^i\,$

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

Take (4) and mutliply by $b$ on the left and (5) and multiply by $a - 1$ on the left and we get

$ba^{i+1}ba^{-i}=bab=a^{i+1}b^2a^{-i}\,$

Now cancelling the $b a - i$ 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 $G$ .