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Question:

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

ord(x) = min{[itex]r \geq 1: x^r = 1[/itex]}

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

My answer: Please verify

If [itex]\theta(x)[/itex] = {[itex]x^r: r \epsilon Z[/itex]} then ord([itex]\theta(x)[/itex]) = 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]" where [G] is an index of H in G, an integer.

So order for any xEG divides order of the group. So ord([itex]\theta(x)[/itex]) = ord(x)

any suggestions or changes please? thnx :)

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# Orders of Groups

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