Suppose that $\phi$ is a homomorphism from a finite group $G$ onto $\overline G$ and that $\overline G$ has an element of order $8$ .Prove that $G$ has an element of order $8$.
Attempt:
I write $e$ to denote the identity of both $G,\overline G$
Let $\phi:G\to \overline G$ be the homomorphism. Let $a_1\in \overline G$ such that $o(a_1)=8$ .Then $\exists a\in G$ such that $\phi(a)=a_1$.
Now $a_1^8=e\implies \phi(a)^8=e\implies \phi(a^8)=e=\phi(e)$.
How to proceed after this?
On
If $f: G_1 → G_2$ is homomorphism and $a∈G_1$ then $o(f(a)) | o(a)$ where $o(a)$ denotes the order of $a$.
Here as, as given that $a_1 ∈ G_2$ has order $8$ that is, $o(a_1) = 8$. As homomorphism is onto so there must be some $a∈G_1$ such that $f(a) = a_1$.
But order $o(f(a)) | o(a)$
$→ o(a_1) | o(a)$
$→ o(a) = 8k$ where k is positive integer. Hence it follows that, $G_1$ has element of order $8$
Take your argument. We know* that $8$ divides the order of $a$, so write $\mathrm{ord}(a) = 8k$ for (the only possible) integer $k \geq 1$. Then the element $a^k \in G$ has order 8. (Certainly $8$ kills $a^k$, and if anything smaller did this would contradict the order of $a$ being $8k$.)
(*) If you didn't know this, it follows from Lagrange. Any homomorphism $\theta: G \to L$ (surjective or not, it doesn't matter) has the property that $\mathrm{ord}(\theta(a))$ divides $\mathrm{ord}(a)$ for each $a \in G$ of finite order.