Let G be a group and a b elementof G Show that a and a1 hav

Let (G, *) be a group and a, b elementof G. Show that a and a^-1 have the same order Show that a and b * a * b^-1 have the same order. Show that a * b and b * a have the same order.

Solution

(i) Let the order of a be n. Then aⁿ = e. Further, e = (aa-1)n = an(a-1)n = e (a-1)n = (a-1)n . Therefore, the order of a-1 n. Let the order od a-1 be m with m < n. Then (a^-1)^m = e . Also, e = e^m = (a^-1a)^m = (a^-1)^m a^m = ea^m = a^m . Thus, a^m = e. However, this is a contradiction. Hence m = n i.e. the order of a^-1 is same as the order of a.

(ii) Let the order of a be n. Then (bab^-1 )ⁿ = bab^-1 · bab^-1 · bab1 · ... · bab^-1 ( n times)

= ba · (b^-1 b) · a · (b^-1 b) · a · (b^-1 b) · ... · (b^-1 b) · ab^-1 = ba · e · a · e · a · e · ... · e · ab^-1 = baⁿ b 1 = beb^-1 = bb^-1 = e.Therefore, the order of bab^-1 n. Let us suppose that the order of bab^-1 is m and that m < n. Then, we must have   e = (bab^-1 )^m = ba^mb^-1 .Therefore, b^-1 eb= b^-1 ba^m b^-1 b = (b^-1b)a^m (b^-1b) or, e = a^m . This implies that the order of a is less than n, which is a contradiction. Hence the order of bab^-1 is n.

(iii) Let the order of ab = n. Then (ab)n =e where e is the identity. Then a^-1(ab)n = a^-1. Then a^-1(ab)n a = a^-1 a = e. Further, a^-1(ab)n a = a^-1 (ababab…)a = a^-1a(ba)ⁿ = (ba)ⁿ =e . Therefore, the order of ba is less than or equal to the order of ab. Let the order of ba be m with m < n. Then (ab)^m = abab….ab(m times) = abab….abe = abab….ab(aa^-1) = a(ba)^m a^-1 = aea^-1 = aa^-1 = e . This is a contradiction. Hence m = n i.e. the order of ba is same as the order of ab.