Let $f:G \rightarrow G'$ be a homomorphism. Let $a \in G$ be such that $o(a)=n$ and $o(f(a))=m$. Show that $f$ is one-one iff $m=n$

Question

Let $f:G \rightarrow G'$ be a homomorphism. Let $a \in G$ be such that $o(a)=n$ and $o(f(a))=m$. Show that $f$ is one-one iff $m=n$

I know that $m|n$ and can prove that $n|m$ which implies that $m=n$. But how to prove converse?

One more doubt:

Why do we need to prove converse when there is an "iff" in the statement?

Edit: Here $o(a)$ means order of 'a'.

Answer 1

The statement of your problem is a little confusing. Is this what you want to say?

Let $f: G \rightarrow G'$ be a group homomorphism. For each $a \in G$, let $n_a$ be the order of $a$, and let $m_a$ be the order of $f(a)$. Then $f$ is one to one if and only if for every $a \in G$, we have $n_a = m_a$.

For your second question, "Why do we need to prove the converse when there is an iff in the statement?" An 'if and only if' is way of writing down two statements at once. That is, you are supposed to prove

(1) If $f$ is one to one, then for every $a \in G$, we have $n_a = m_a$

and

(2) If for every $a \in G$, we have $n_a = m_a$, then $f$ is one to one.

So you are definitely going to need to prove the converse of something when you are asked to prove an if and only if. The statements (1) and (2) are converses of each other.