let f: G-->G\' be a homomorphism with kernel H. assme that G is finite. a) show
ID: 3107935 • Letter: L
Question
let f: G-->G' be a homomorphism with kernel H. assme that G is finite.a) show that order of G= (order of image of f)(order of H)
b) suppose that G, G' are finite groups and that the orders of G and G are relatively prime. prove that f is trivial, that is, f(G)=e', the unite element of G'
Explanation / Answer
First we need a lemma: Suppose f(z) = f(w), then zH = wH (this is notation for cosets of the kernel). proof: f(z) = f(w) ==> ([f(w)]^-1)f(z) = e ==> f([w^-1]z) = e (since f is a homomorphism). Now by defn [w^-1]z is in H, thus [w^-1]zH = H and it follows zH = wH a) proof: Let y be in the f(G) (the image of f). Then there is some x in G such that f(x) = y. We want to look a f^-1(y), that is every element in G that is mapped to y by f. It is not hard to verify f^-1(y) = xH (this follows from the above lemma, write back if its not clear), thus the order of f^-1(y) is the same as the order of xH which is clearly the same as the order of H. Now we summing over every y in f(G) we get the desired result. Note if you already know the first isomorphism theorem for groups this is immediate. The 1st iso thm say G/H is isomorphic to f(G), since everything is finite the order of G/H is just (order of G)/(order of H), multiplying by (order H) on both sides gives the desired result. b) Part a implies the order of f(G) divides the order of G. We also know f(G) is a subgroup of G` (since f is a homomorphism), thus (order f(G)) divides two relatively prime integers so we must have (order f(G)) = 1, and the only subgroup of order 1 is the trivial subgroup.
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