Homomorphism of groups and order of elements

Question

This is what I have to show:

Let $f: G\to H$ be a homomorphism and let $p$ be prime. If range of $f$ has an element of order $p$, then $G$ has an element of order $p$.

I start off by proving the lemma:

Lemma. For each element $a\in G$, the order of $f(a)$ is a divisor of order of $a$.

Proof. Suppose $o(a)=n$. Then $a^{n}=e$. Applying $f$ both sides, we have $f(a^n)=[f(a)]^n = e$. Thus, $o(f(a))$ divides $n=o(a)$. QED

And this is how I start the proof of the question:

Let $a\in $ range of $f$ with $o(a)=p$. Thus, $a=f(b)$ for some $b \in G$. From the lemma, we know that $o(f(b))=o(a)=p$ divides $o(b)$. Now, it remains to show $o(b)$ divides $p$ and I'll be done. However this is the part I get stuck.

I also know range of $f$ is a subgroup of $H$. However I don't see how that information helps me.

Can I get some hints?

Answer 1

First you need $G$ to be finite. Try to think of an example. Suppose that $o(a)$ = 15. This means $a^{15} = e$ Can you now find an element $b$ (built from $a$) such that $b^3 = e$? Once you understand the example, try to generalize.

Answer 2

Once you know that $p$ divides the order of $b$ you are done. The element $b^{(o(b)/p)}$ has order $p$.

Answer 3

hint: since $p \mid o(a)$, we have that

$a^{(mp)}=e$, so $(a^m)^p=e$.