**Theorem**: Let G be a group with identity e. If a ∊ G and O(a) = m, then for some positive integer n, a^{n} = e if and only if m | n (m divides n).

Proof: O(a) = m

⇒ m is the least positive integer such that a^{m} = e. Hence a^{n} = e.

⇒ n ≥ m.

By division algorithm we can write n = h m +r,

## where h is an integer and r is a non negative integer less than m.

Now e = a^{n}

= a^{hm+r}

^{ } = a^{hm} a^{r}

= (a^{m})^{h} a^{r}

= e^{h} a^{r}

= e a^{r}

= a^{r}

We have a^{r} = e with 0 ≤ r < m.

Since m is the smallest positive integer such that a^{m} = e.

We must have r = 0.

Thus n = hm which implies that m is a factor of n.

i.e., m | n.

Conversely, let m | n. Then n hm for some integer h,

and so a^{n} = a^{hm}

= (a^{m})^{h}

= e^{h}

= e.

Thus the theorem is proved.