Hi i have completed the answer to this question. Just need your verification on whether it's completely correct or not:

Question:

If G is a group and xEG we define the order ord(x) by:

ord(x) = min{$\displaystyle r \geq 1: x^r = 1$}

If $\displaystyle \theta$: G --> H is an injective group homomorphism show that, for each xEG, ord($\displaystyle \theta(x)$) = ord(x)

My answer: Please verify

If $\displaystyle \theta(x)$ = {$\displaystyle x^r: r \epsilon Z$} then ord($\displaystyle \theta(x)$) = ord(x).

For any integer r, we have x^r = e (or 1) if and only if ord(x) divides r.

In general the order of any subgroup of G divides the order of G. If H is a subgroup of G then "ord (G) / ord(H) = [G:H]" where [G:H] is an index of H in G, an integer.

So order for any xEG divides order of the group. So ord($\displaystyle \theta(x)$) = ord(x)

any suggestions or changes please? thnx