Let f:G to H be any group homomorphism, that is, f(x)f(y) is always f(xy). It will follow that also f(x^{-1})=(f(x))^{-1}. Let K be its kernel, the set of all elements of G such that f(x)=e. To do this, let a and b be in K and let g be any element of G. Being in K means f(a)=f(b)=e. Make calculations to prove f(ab)=e f(a^{-1})=e f(gag^{-1})=e

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