Let G be a finite group and let f :Q→G be a group homomorphism of
additive group Q into G. Then show that f(x)=e for all x∈Q.
additive group Q into G. Then show that f(x)=e for all x∈Q.
-
Let f :Q→G be a group homomorphism, where |G| = n for some positive integer n.
So, Lagrange's Theorem implies that g^n = e for all g in G.
Given any x in Q, we have
f(x) = f(n * x/n)
.....= [f(x/n)]^n, since f is a homomorphism
.....= e, since f(x/n) is in G.
Hence, f is the trivial homomorphism.
I hope this helps!
--------------
P.S.: f(0) = e was not needed; all that says is that the identity maps to the identity under a group homomorphism.
The other part was a typo. from an earlier version of the proof which is now fixed.
So, Lagrange's Theorem implies that g^n = e for all g in G.
Given any x in Q, we have
f(x) = f(n * x/n)
.....= [f(x/n)]^n, since f is a homomorphism
.....= e, since f(x/n) is in G.
Hence, f is the trivial homomorphism.
I hope this helps!
--------------
P.S.: f(0) = e was not needed; all that says is that the identity maps to the identity under a group homomorphism.
The other part was a typo. from an earlier version of the proof which is now fixed.