If ord(x)=a then .
Now suppose that .
Then
(injective)
Do you see the contradiction?
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{ }
If : G --> H is an injective group homomorphism show that, for each xEG, ord( ) = ord(x)
My answer: Please verify
If = { } then ord( ) = 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( ) = ord(x)
any suggestions or changes please? thnx
yes i think i see the contradiction.
i believe to do this question we have to show 2 things:
i)
ii) .
therefore the contradiction seems to arise from the fact that if , then which is not what statement (ii) says.
am i correct in this assumption? any ideas on how to deal with this?