Answer :
The proof of (g/n)/(g/m) isomorphic to g/m from the first isomorphism theorem is as shown below.
Let a group g with normal subgroups m and n. Furthermore, n is a subset of m.
Proof: We have to prove (g/n)/(m/n) ≅ g/m (1)
g/m and g/n are valid because m and n are normal in g. As normality satisfies the intermediate subgroup condition, n is normal in m. because normality is image-closed m/n is a normal subgroup in g/n. Under the quotient map by n, the normal subgroup m of g is sent to a normal subgroup m/n of g/n. Thus the L.H.S of the statement to prove makes sense.
To describe the isomorphism from the left side to the right side
φ: g/n \to g/m
ψ(gn) = gm
So, a coset of n is taken by the map and it gives the corresponding coset of m. This is well-defined, because if h∈ n, thus h∈ m, hence (gh)m = g(hm) = gm.
The map is a homomorphism. So we observe that it maps the identity element to the identity element, keeps group multiplication preserved, and also preserves the inverse map.
Also, the map is surjective because any coset gm is present as the image of gn under ψ.
Lastly, we need to find the kernel of the map which is given by the set of gn such that gm = m. This is those cosets of n that are in m, and this is the same as the coset space m/n. Thus the kernel of the map is exactly g/m. Thus, the surjective homomorphism ψ: g/n → g/m has kernel m/n. And now by the first isomorphism theorem
(g/n)/(m/n)≅g/m where ≅ shows isomorphism
another problem to prove whether f is an isomorphism:
brainly.com/question/19865639
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