Let f: G->H be a surjective homomorphism of groups with kernel K and let M be a subgroup of H?

Question

(a) Prove that there is a subgroup N of G such that K is a subset of N is a subset of G and N/K is isomorphic to M.
(b) If M is normal in H, prove that N is normal in G and G/N is isomorphic to H/M

Answer 1

I emailed you complete and worked out solutions to your problems...I was having an issue typing the symbols which are required for the proof on here, and I thought that the solutions would be better understood with the proper symbol usage. Let me know if you have questions, I would be happy to explain something further.

Answer 2

a) Kernel K < G is equivalent to saying the K is the preimage of 1H (identity of H). The cosets of K are exactly what get sent to each element of M. That is, given m in M, let p in G be a preimage of m. Then the entire coset pK gets sent to m. These cosets constitute N, which is a group by including the identity (always in K) and closure comes from M, and by construction N/K is isomorphic to M.

b) Using the conjugation definition of normal subgroup, the normality of H carries easily back to N. Looking at cosets gives the quotient isomorphism.

I know these are on the intuitive side; details come from the homomorphism theorems and carefully quoting the relevant definitions. If the concepts are still unclear, look at some concrete examples, say D6 to Z3.

Answer 3

(a) Let N = {x∈G | f(x) ∈ M} (ie. N = f^-1 (M)). Since

f(k) = e∈M ∀ k∈K

(where 'e' will throughout denote the identity element of H or G according to context), we have that K⊆N. To show that N is a subgroup of G, observe that for x, y∈N:

f(xy) = f(x)f(y)∈M and f(x^-1)=f(x)^-1

since f is a group homomorphism and M is a group, so xy∈N and x^-1 ∈M.

N/K is isomorphic to M via the isomorphism g, where g is given by

g(Kx) = f(x) for each Kx∈N/K;

this is well defined because Kx = Ky means xy^-1∈K, and is surjective by our definition of N and is a group homomorphism because f is a homomorphism. To show that g is injective, observe that for x, y ∈ N

g(Kx) = g(Ky) ⇒ f(x) = f(y) ⇒ f(x)f(y)^-1 = e ⇒ f(xy^-1) = e

⇒ xy^-1∈K ⇒ Kxy^-1 = K ⇒ Kx=Ky.

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(b) Let x∈G. We will show that xN⊆Nx. Let y∈N. Then

f(xy) = f(x)f(y) = f(z)f(x)

(for some z∈N, because f(x)M = Mf(x) (M is normal in H) and f maps N onto M)

= f(zx), ie. f(xyx^-1) = f(z)∈M. From this it follows that xyx^-1∈N (by definition of N), and so xy∈Nx.

Nx⊆xN follows by symmetry (ie. a tedious retread of the same sort of argument just shown). Hence Nx = xN for any given x∈G.

It remains to prove that G/N is isomorphic to H/M. Define a map h from G/N to H/M by

h(Nx) = Mf(x).

This is a well defined since Nx = Ny means that xy^-1∈M, and is a homomorphism onto H/M since f is a homomorphism onto H. It remains to show that h is injective. Suppose we have x, y∈G with Mf(x) = Mf(y). Then

M = f(x)^-1Mf(y) = f(x^-1)Mf(y) = Mf(x^-1)f(y)

(because M is normal)

= Mf(x^-1y).

Thus f(x^-1y)∈M, which implies that x^-1y∈N; hence y∈xN = Nx (because N is normal, as already shown). QED.

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You may wish to flesh out some of the steps here.

Answer 4

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