Let $G$ and $H$ be groups, and let $f: G \rightarrow H$ be a group homomorphism. We are asked to prove that if $f$ is injective, then for any element $a \in G$, the order of $a$ in $G$ (denoted $\text{ord}(a)$) equals the order of $f(a)$ in $H$ (denoted $\text{ord}(f(a))$).

Step-by-Step Proof:

1. Definition of Order of an Element:

   • The order of an element $a \in G$, denoted $\text{ord}(a)$, is the smallest positive integer $n$ such that: $a^n = e_G$ where $e_G$ is the identity element in $G$.
   • Similarly, the order of an element $f(a) \in H$, denoted $\text{ord}(f(a))$, is the smallest positive integer $m$ such that: $f(a)^m = e_H$ where $e_H$ is the identity element in $H$.

2. Homomorphism Property: Since $f$ is a homomorphism, we know that: $f(a^n) = f(a)^n$ for any integer $n$. In particular, for $n = \text{ord}(a)$, we have: $f(a^{\text{ord}(a)}) = f(e_G) = e_H.$ Thus: $f(a)^{\text{ord}(a)} = e_H.$

3. Injectivity of $f$: Suppose $f$ is injective. By the definition of injectivity, we have that if $f(x) = f(y)$, then $x = y$. This implies that if $f(a)^{\text{ord}(a)} = e_H$, then: $f(a^{\text{ord}(a)}) = f(e_G) = e_H,$ and because $f$ is injective, the element $a^{\text{ord}(a)}$ must be the identity in $G$. Therefore: $a^{\text{ord}(a)} = e_G.$

4. Minimality of $\text{ord}(a)$: Since $f(a)^{\text{ord}(a)} = e_H$, we know that $\text{ord}(f(a)) \leq \text{ord}(a)$. Now, assume for contradiction that $\text{ord}(f(a)) < \text{ord}(a)$. This would mean that there exists a smaller integer $m$ such that: $f(a)^m = e_H \quad \text{with} \quad m < \text{ord}(a).$ By the homomorphism property: $f(a^m) = f(a)^m = e_H,$ and since $f$ is injective, we must have $a^m = e_G$, which contradicts the fact that $\text{ord}(a)$ is the smallest such integer.

   Therefore, we conclude that $\text{ord}(f(a)) = \text{ord}(a)$.

Conclusion:

Thus, we have shown that if $f$ is injective, then for any $a \in G$, the order of $a$ in $G$ is equal to the order of $f(a)$ in $H$. Therefore: $\text{ord}(a) = \text{ord}(f(a)).$

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Would you like further details or clarifications on any part of the proof? Here are some related questions to consider:

1. What does it mean for a group homomorphism to be injective, and how does it relate to kernel?
2. How would this result change if $f$ were surjective instead of injective?
3. Can you provide an example of an injective homomorphism between two groups and compute the orders?
4. What is the general relationship between the kernel of a homomorphism and the order of elements in the domain and codomain?
5. How does the order of an element relate to powers of that element in group theory?

Tip: When proving properties of homomorphisms, always consider how the kernel and image of the homomorphism influence the structure of the groups involved.