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Home / Modern Abstract Algebra / The order of any element in G is the same as that of its inverse in G.

The order of any element in G is the same as that of its inverse in G.

Theorem: If G is a group, then (i) for every a ∊ G, 0(a) = 0(a-1).

i.e., the order of any element in G is the same as that of its inverse in G.

(ii) for any a, x ∊ G.

0(a) = 0(x-1ax)

Proof: (i) Let 0(a) = r and 0(a-1) = s.

Then ar = e and (a-1)s = e, where e is the identity in G. Let r ≠ s. Then either r>s or r<s.

Now, if r>s ⟹0<r-s<r

And then ar-s =ar a-s = ar (a-1)s = e e = e

This contradicts the assumptions that 0(a-1) = s.

Again, r < s ⟹0 <r – s <s

And then (a-1)s-r = (a-1)s (a-1)-r = (a-1)s ar = e e = e.

This contradicts the assumption the that

0(a-1) =s.

Hence r = s, i.e., 0(a) = 0(a-1)

If 0(a) is infinite then also 0(a-1) is infinite. For if 0(a-1) = m, a positive integer, then (a-1)m = e ⟹ (am)-1 = e ⟹am = e-1 = e

0(a) is finite which is a contradiction.

(ii) Let b = x-1 a x

Then b2 = (x-1 a x)2 = (x-1 a x) (x-1 a x) = x-1 a x x-1 a x = x-1 a e ax

= x-1 a ax = x-1 a2x

b3 = b2b = (x-1 a2x)( x-1 a x) = x-1 a2 x x-1 a x = x-1 a2 e a x = x-1 a3 x

and generally, bn = x-1 an x for any positive integer n. Thus if an = e, then

bn = x-1ex = e.

Conversely if bn = e, then x-1 an x = e

⟹ an x = xe = x ⟹ a n = xx-1=e

Hence 0(a) = 0(b). i.e., 0(a) = 0(x-1ax).   (proved )

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