Alg7.17

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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.


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