Let $\varphi: G \rightarrow H$ be a group homomorphism, and $a, b \in G$ such that $|a| = |b|$, is $|\varphi(a)|=|\varphi(b)|$?

Question

Let $\varphi: G \rightarrow H$ be a group homomorphism, and $a, b \in G$ such that $|a| = |b|$, is $|\varphi(a)|=|\varphi(b)|$?

Let $k=|a| = |b|$, then $|\varphi(a)|$ and $|\varphi(b)|$ should be divisors of $k$.

I don't know where to go from here.

Answer 1

Here's a simple counterexample:

Let $C_2=\{1,g\}$ be a cyclic group of order 2 with generator $g$, let $G=C_2 \times C_2$, $H = C_2$, and let $\varphi : G\to H$ be defined by $(a,b)\mapsto a$.

Now we have $$|(g,1)|=2=|(1,g)|$$ but $$|\varphi((g,1))|=|g|=2\ne1=|1|=|\varphi((1,g))|$$