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This proof (one minor detail is missing, really) of the first isomorphism theorem uses the symbols in my writeup there without repetition of the formulation, so you may want to read that first.

Let K=Ker f. If g is in G, and k is in K, then f(k-1hk) = f(k-1) * 1 * f(k) = 1, and since we may replace g by g-1 in the above, this shows K is normal.

Note that for any g in G, f maps all of gK to the same element of H; therefore the induced function f:G/K->H defined by f(gK) = f(g) is well defined. It's easy to verify it is indeed a homomorphism.

It's a monomorphism, since by definition f(gK) = 1 iff f(g) = 1. And it's onto Im f by definition. Hence it's an isomorphism, which proves the theorem.

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