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I suspect it is quite simple and works by contradiction, but I can't see why for a group homomorphism f, ker(f) = {e} ==> f is injective. Any help?
I hadn't realized that.matt grime said:why would you want ax=xa anyway? that is not the condition of normality of a subgroup. H is normal if for gx in gH there is an element y of H such that yg=gx, there is no requirement that x=y.
May I have a hint?matt grime said:did you redo the proof of the last problem to be a one line proof NOT by contradiction? it would be helpful to your understanding to do it.
Sure you have. The basic group operation is often called multiplication unless otherwise specified. But if the use of the word "multiplication" confuses you, ignore that word and replace that part of my post with:quasar987 said:Ooh, I see .
(I haven't seen multiplication yet AKG)
matt grime said:i think it's quite straight forward, if H is the inerse image of H', then gx in H...
matt grime said:...implies [tex]f(gx) \in f(gH)=f(g)H'=H'f(g)[/tex] which implies that there is a y in H such that f(yg)=f(xg)
matt grime said:...so that, pulling back by f gx is in Hg too.
If a is an element of a group G, there is always a homomorphism from Z to G which sends 1 to a. When is there a homomorphism from [itex]\mathbb{Z}_n[/itex] to G which sends [1] to a? What are the homomorphisms from [itex]\mathbb{Z}_2[/itex] to [itex]\mathbb{Z}_6[/itex]?
Suppose G is a group and g is an element of G, g [itex]\neq[/itex] e. Under what conditions on g is there a homomorphism f : Z_7 --> G with
f([1]) = g ?