Suppose that G is a simple group and f: G rightarrow H is a surjective homomorph

Question

Suppose that G is a simple group and f: G rightarrow H is a surjective homomorphism of groups. Prove that either f is an isomorphism or H = . Let G be an abelian group. Show that K = {a G ||a| le 2} is a subgroup of G. Show that H = {x^2 | x G} is a subgroup of G. Prove that G/K H. If N is a normal subgroup of a group G and T is a subgroup of G/N, show that H = {a G| Na T} is a subgroup of G. If k|n and f: U_n rightarrow U_k is given by f([x]_n) = [x]_k, show that f is a homomorphism and find its kernel.

Explanation / Answer

f: G to H

Given that f is a surjective homomorphism

If H=(e) this is trivially true.

Because if a and b are two elements in G

then f(a*b) = f(a)*f(b) = e*e =e

If H consists of other elements than e, say c

then c must be its own inverse

If f(a) =c, f(b) = e

then f(a*b) = c

Since f is an homomorphism that admits an inverse.

So f is a isomorphism.

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