Let G be a group and x belongs to G,
Define ord(x) = min{r >= 1 : x^r = 1}
If f: G map to H is an injective group homomorphism. Show that, for
each x in G, ord(f(x)) = ord(x).
I think I come up with the approach is that , I need to show that
ord(f(x)) divides r and hence the result is proved.
Or: I can somehow show that ord(f(x)) = min{ r >= 1 : (f(x))^r = 1} which is by the definition.
Can you tell me whether the above approaches are right or wrong, if they are wrong then show me how to solve this question please?
[Thus, but then .
Say that there is with but then .
To ThePerfectHacker, thanks for your answer, but there are some points
I am not sure I understand, how come you got: . And finally when you concluded that . How can you imply this result?
Thank you again for your precious time
because is a homomorphism. Since and is a group homomorphism, which means it maps identity to identity. ThusI am not sure I understand, how come you got:
He assumed that \phi is injective (or one-one). And , thus injectivity forces .. And finally when you concluded that . How can you imply this result?