**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^{-1}ax)

**Proof:** (i) Let 0(a) = r and 0(a^{-1}) = s.

Then a^{r} = 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 a^{r-s }=a^{r} a^{-s} = a^{r} (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} a^{r} = 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 ⟹ (a^{m})^{-1} = e ⟹a^{m} = e^{-1} = e

⟹*0(a) is finite which is a contradiction.*

(ii) Let b = x^{-1} a x

Then b^{2} = (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} a^{2}x

b^{3} = b^{2}b = (x^{-1} a^{2}x)( x^{-1} a x) = x^{-1} a^{2 }x x^{-1} a x = x^{-1} a^{2} e a x = x^{-1} a^{3} x

and generally, b^{n} = x^{-1} a^{n }x for any positive integer n. Thus if a^{n} = e, then

b^{n} = x^{-1}ex = e.

Conversely if b^{n} = e, then x^{-1} a^{n} x = e

⟹ a^{n} x = xe = x ⟹ a ^{n} = xx^{-1}=e

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