Let f : G ! G′ be an onto homomorphism. For H′ a subgroup of G′, define
H = {x 2 G|f(x) 2 H′}.
Then show that
(i) H is a subgroup of G
To show that H is a subgroup of G, we need to show that it satisfies the three defining properties of a subgroup:
1. The identity element e is in H: Since f is onto, there exists an element x in G such that f(x) = e′, where e′ is the identity element of G′. Therefore, x ∈ H since f(x) = e′ is also in H′. Thus, the identity element of G is in H.
2. H is closed under the binary operation of G: Let a, b ∈ H. Then f(a), f(b) ∈ H′ by definition. Since H′ is a subgroup of G′, f(a)f(b) ∈ H′. But f is a homomorphism, so f(ab) = f(a)f(b). Therefore, ab ∈ H, and H is closed under the binary operation of G.
3. H is closed under inverses: Let a ∈ H. Then f(a) ∈ H′, so (f(a))−1 ∈ H′ since H′ is a subgroup of G′. But f is a homomorphism, so f(a−1) = (f(a))−1. Therefore, a−1 ∈ H, and H is closed under inverses.
Since H satisfies all three properties of a subgroup, H is a subgroup of G.
(ii) If K′ is another subgroup of G′ such that K′ ⊆ H′, then K ⊆ H:
Let k ∈ K. Then f(k) ∈ K′ since f is onto and K′ ⊆ H′. But K′ ⊆ H′ implies that f(k) ∈ H′. Therefore, k ∈ H by definition. Since k was arbitrary, every element of K is also in H, so K ⊆ H.
Let f : G ! G′ be an onto homomorphism. For H′ a subgroup of G′, define
H = {x 2 G|f(x) 2 H′}.
Then show that
(i) H
(ii) H′ is a normal subgroup of G′ if H is normal in G.
(i) Incomplete. Please provide the missing context.
(ii) Let H be a normal subgroup of G. We need to show that H′ is a normal subgroup of G′. Let g′ ∈ G′ and h′ ∈ H′ be arbitrary elements. We need to show that g′h′g′^-1 ∈ H′.
Since f is onto, there exists an element g in G such that f(g) = g′. Let h ∈ H be the unique element such that f(h) = h′ (which exists because f is onto). Then we have:
f(g'h'g^-1) = f(g')f(h')f(g)^-1 = g'f(h')g^-1
where we have used the fact that f is a homomorphism. Since H is normal in G, we have g'hg^-1 ∈ H for all g ∈ G and h ∈ H. In particular, we have g'hg^-1 ∈ H for g = g' and h = h'. Therefore, g'f(h')g^-1 = f(g'hg^-1) ∈ H', since H is defined as the set of elements x in G such that f(x) ∈ H'. So we have shown that g'h'g'^-1 ∈ H', which means that H' is a normal subgroup of G'.
To show that H is a subgroup of G, we need to prove three things: closure, identity, and inverse.
(i) Closure: Let a, b be two elements in H. This means that f(a), f(b) are both in H'. Since H' is a subgroup of G', it is closed under the group operation. Therefore, f(a) * f(b) is in H'. Now, since f is onto, there exist elements x, y in G such that f(x) = f(a) and f(y) = f(b). Thus, we have x * y = a * b since f is a homomorphism. This shows that a * b is in H.
(ii) Identity: Since f is onto, there exists an element e' in G' such that f(e) = e'. Let's show that e is the identity element in H. Let a be any element in H. We have f(a) in H', so f(a) * e' = f(a), since e' is the identity element in H'. Now, since f is onto, there exists an element x in G such that f(x) = f(a). Therefore, we have x * e = a, showing that e is the identity element in H.
(iii) Inverse: Let a be an element in H. This means that f(a) is in H'. Since H' is a subgroup, it contains the inverse of f(a). Let's call it (f(a))^-1. Now, since f is onto, there exists an element x in G such that f(x) = f(a). Therefore, we have (f(a))^-1 = f(a)^-1 = f(x)^-1 = f(x^-1), since f is a homomorphism. This shows that x^-1 is the inverse of a in H.
Since H satisfies closure, identity, and inverse, it is a subgroup of G.