Abstract algebra: homomorphism and an element of a prime order Let me ask Prob.F6 in Pinter's algebra book.
'Let $p$ be a prime and $f$ be a homomorphism from $G$ to $H$ where $G$ and $H$ are groups.
 If range of $f$ has an element of order $p$, then $G$ has an element of order $p$.' 
If $f$ is injective, life gets easier. But without the injectivity of the homomorphism, i feel lost... Would somebody help please?
 A: It is true if $G$ is a group in which all elements have finite order (besides finite groups, you have the example of $\mathbf Q/\mathbf Z$)
Indeed, let $a\in G$ an element of order $n$ such that $f(a)$ has order $p$.Then $p$ is a divisor of $n$: $n=pm$  for some $m>0$, and $b=a^m$ has order $p$.
A: If $G$ is not assumed to be finite (or at least that every element of $G$ has finite order), we can't conclude, because the canonical homomorphism $\mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}$ provides a counterexample: the range has an element of order $p$, but $\mathbb{Z}$ has no element of order $p$.
So, assume $x$ has finite order $n$ and that $f(x)$ has order $p$. We can define a homomorphism $f_x\colon\langle x\rangle\to H$ by restricting $f$. By the homomorphism theorem, we can say that $p\mid n$: indeed the order of the range of $f_x$ is always a divisor of the order of the domain, because the range is isomorphic to the quotient of the domain modulo the kernel. By assumption, the range of $f_x$ has order $p$, because it is $\langle f(x)\rangle$.
Thus the group $\langle x\rangle$ has order divisible by $p$. Cauchy's theorem allows us to finish.
A: Surely false consider the additive group $G=\mathbb{Z}$ and $H= \mathbb{Z}_5$, and the homomorphism is the map modulo 5. Now the image has an element of order 5 but $G$ does not have any element of order 5.
