(a) e_H = o(e_G)

This shows that o takes the identity element of G to the identity element of H.

(b) By the principle of mathematical induction, the statement o(g^n) = (o(g))^n holds for all n ∈ Z.

(c) we have shown that o(g^[g]) = e_H, which implies that [g] divides [g^[g]].

(d) Since Kero is closed under the **group operation**, contains the identity element, and contains inverses, it is a subgroup of G.

(e) Combining both **directions**, we have proven that o(a) = o(b) if and only if aKero = bKero.

(f) Combining both **inclusions**, we have gKero = o^(-1)(h) = {r ∈ G : o(r) = h}.

(a) To show that o takes the **identity **of G to the identity of H, we need to prove that o(e_G) = e_H, where e_G is the identity element of G and e_H is the identity element of H.

Since o is a **homomorphism**, it preserves the group operation. Therefore, we have:

o(e_G) = o(e_G * e_G)

Since e_G is the identity element, e_G * e_G = e_G. Thus:

o(e_G) = o(e_G * e_G) = o(e_G) * o(e_G)

Now, let's multiply both sides by the **inverse **of o(e_G):

o(e_G) * o(e_G)^-1 = o(e_G) * o(e_G) * o(e_G)^-1

Simplifying:

e_H = o(e_G)

This shows that o takes the identity element of G to the identity element of H.

(b) To prove that o(g^n) = (o(g))^n for all n ∈ Z, we can use induction.

Base case: For n = 0, we have g^0 = e_G, and we know that o(e_G) = e_H (as shown in part (a)). Therefore, (o(g))^0 = e_H, and o(g^0) = e_H, which satisfies the equation.

Inductive step: Assume that o(g^n) = (o(g))^n holds for some integer k. We want to show that it also holds for k + 1.

We have:

o(g^(k+1)) = o(g^k * g)

Using the **homomorphism property** of o, we can write:

o(g^(k+1)) = o(g^k) * o(g)

By the **induction hypothesis**, o(g^k) = (o(g))^k. Substituting this in the equation, we get:

o(g^(k+1)) = (o(g))^k * o(g)

Now, using the property of exponentiation, we have:

(o(g))^k * o(g) = (o(g))^k * (o(g))^1 = (o(g))^(k+1)

Therefore, we have shown that o(g^(k+1)) = (o(g))^(k+1), which completes the induction step.

By the principle of mathematical induction, the statement o(g^n) = (o(g))^n holds for all n ∈ Z.

(c) If g is finite, let [g] denote the order of g. The order of an element g is defined as the smallest **positive integer** n such that g^n = e_G, the identity element of G.

Using the homomorphism property, we have:

o(g^[g]) = o(g)^[g] = (o(g))^([g])

Since o(g) has finite order, let's say m. Then we have:

(o(g))^([g]) = (o(g))^m = o(g^m) = o(e_G) = e_H

Therefore, we have shown that o(g^[g]) = e_H, which implies that [g] divides [g^[g]].

(d) To prove that Kero = {g ∈ G : o(g) = e_H} is a subgroup of G, we need to show that it is closed under the group operation, contains the identity element, and contains **inverses**.

Closure under the group operation: Let a, b ∈ Kero. This means o(a) = o(b) = e_H. Since o is a homomorphism, we have:

o(a * b) = o(a) * o(b) = e_H * e_H = e_H

Therefore, a * b ∈ Kero, and Kero is closed under the group operation.

Identity element: Since o is a homomorphism, it maps the identity element of G (e_G) to the identity element of H (e_H). Therefore, e_G ∈ Kero, and Kero contains the identity element.

Inverses: Let a ∈ Kero. This means o(a) = e_H. Since o is a homomorphism, it preserves inverses. Therefore, we have:

o(a^-1) = (o(a))^-1 = (e_H)^-1 = e_H

Thus, a^-1 ∈ Kero, and Kero contains inverses.

Since Kero is closed under the **group operation**, contains the identity element, and contains inverses, it is a subgroup of G.

(e) To prove the statement "o(a) = o(b) if and only if aKero = bKero":

Forward direction: Suppose o(a) = o(b). This means that a and b have the same image under the homomorphism o, which is e_H. Therefore, o(a) = o(b) = e_H. By the definition of Kero, we have a ∈ Kero and b ∈ Kero. Thus, aKero = bKero.

Backward direction: Suppose aKero = bKero. This means that a and b belong to the same coset of Kero. By the definition of cosets, this implies that a * x = b for some x ∈ Kero. Since x ∈ Kero, we have o(x) = e_H. Applying the homomorphism property, we get:

o(a * x) = o(a) * o(x) = o(a) * e_H = o(a)

Similarly, o(b) = o(b) * e_H = o(b * x). Since a * x = b, we have o(a * x) = o(b * x). Therefore, o(a) = o(b).

Combining both **directions**, we have proven that o(a) = o(b) if and only if aKero = bKero.

(f) Suppose o(g) = h. We want to show that o^(-1)(h) = {r ∈ G : o(r) = h} = gKero.

First, let's show that gKero ⊆ o^(-1)(h). Suppose r ∈ gKero. This means that r = gk for some k ∈ Kero. Applying the homomorphism property, we have:

o(r) = o(gk) = o(g) * o(k) = h * e_H = h

Therefore, r ∈ o^(-1)(h), and gKero ⊆ o^(-1)(h).

Next, let's show that o^(-1)(h) ⊆ gKero. Suppose r ∈ o^(-1)(h). This means o(r) = h. Applying the homomorphism property in reverse, we have:

o(g^-1 * r) = o(g^-1) * o(r) = o(g^-1) * h

Since o(g) = h, we have:

o(g^-1) * h = (h)^-1 * h = e_H

This shows that g^-1 * r ∈ Kero. Therefore, r ∈ gKero, and o^(-1)(h) ⊆ gKero.

Combining both **inclusions**, we have gKero = o^(-1)(h) = {r ∈ G : o(r) = h}.

This completes the proof.

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