Question #17443

Prove that in any group G, orders of elements ab and ba are equal.

Expert's answer

Suppose that ab has order n.

That is, suppose that n is the smallest positive integer such that (ab)^n =e, were e is the identity element of G.

Notice that

(ab)(ab)^(n-1) = (ab)^n = e,

so by uniqueness of inverseses,

(ab)^(n-1) = b^(-1)a^(-1)

Then:

(ba)^n = (ba)(ba) ... (ba)

= b(ab)(ab) ....(ab)a

= b(ab)^(n-1)a

= b( b^(-1)a^(-1) )a

= e

This shows that (ba)^n = e. Furthermore, n is the smallest integer for which this equality holds, since if there were positive m < n such that (ba)^m = e, then a symmetric argument would show that (ab)^m=e, contradicting our definition of n as the order of ab.

Thus n is the order of both elements ab and ba.

That is, suppose that n is the smallest positive integer such that (ab)^n =e, were e is the identity element of G.

Notice that

(ab)(ab)^(n-1) = (ab)^n = e,

so by uniqueness of inverseses,

(ab)^(n-1) = b^(-1)a^(-1)

Then:

(ba)^n = (ba)(ba) ... (ba)

= b(ab)(ab) ....(ab)a

= b(ab)^(n-1)a

= b( b^(-1)a^(-1) )a

= e

This shows that (ba)^n = e. Furthermore, n is the smallest integer for which this equality holds, since if there were positive m < n such that (ba)^m = e, then a symmetric argument would show that (ab)^m=e, contradicting our definition of n as the order of ab.

Thus n is the order of both elements ab and ba.

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