Homomorphism of Groups and divisibility of orders I've been studying for my exams and bumped across this one particular question that I've been having a tad difficulty on:
Suppose that $\phi$ is a homomorphism from a finite group $G$ onto $H$ and that $H$ has an element of order $x$. Prove that $G$ has an element of order $x$ as well.
I tried emailing my prof and I was told to go about doing it first on a particular case when $x = 8$. 
I made use of the fact that $\phi$ is onto so $\forall y \in H, \exists x \in G$ such that $\phi (x) = y$. So pick a particular $y$ such that $|y| = 8$ so we have $\phi(x)^8 = \phi(x^8) = y^8 = e$. Then by property of homomorphism we have that the order of $x$ is divisible by 8. Furthermore we have $x^8 \in \ker \phi$. This was about as far as I got and was hoping I can get help to finish this and thus the general case as well.
 A: You are indeed almost there. You know that $x^8\in\mathrm{ker}(\phi)$, so $\mathrm{ker}(\phi)\cap\langle x\rangle$ is nontrivial, and a subgroup of $\langle x\rangle$. In fact, because you are assuming that the order of $y$ is exactly $8$, then no smaller power of $x$ may lie in $\mathrm{ker}(\phi)$. So that means that $\mathrm{ker}(\phi)\cap \langle x\rangle = \langle x^8\rangle$. 
Now, consider the subgroup $\langle x\rangle$. This is cyclic of order $n$ for some $n$. Since $\langle x^8\rangle$ is a subgroup, its index in $\langle x\rangle$ is a divisor of $n$. What's more: $8$ is the smallest power of $x$ that lies in $\langle x^8\rangle$; while this is not true for arbitrary cyclic groups (for example, if the order of $x$ were $12$, then $x^4 = x^{12}x^4 = x^{16}=(x^8)^2$ would also be in $\langle x^8\rangle$), here we know that the smallest power of $x$ that lies in $\mathrm{ker}(\phi)$ is $x^8$, and that $\langle x^8\rangle$ equals the intersection of $\langle x\rangle$ with $\mathrm{ker}(\phi)$. That means that the index of $\langle x^8\rangle$ in $\langle x\rangle$ is exactly $8$, and therefore that the order of $\langle x\rangle$ is a multiple of $8$. Since $n$ is the order of $x$, then the order of $x$ is a multiple of $8$, say $n=8k$. Then what is the order of $x^k$?
Now repeat the argument for the general case.

Added. In fact, you don't need to go that far down. You had already figured out that the order of $x$ is a multiple of $8$ (a simple way to see this: if the order of $x$ is $n$, and $\phi(x)=y$ has order $8$, then $y^n = \phi(x)^n = \phi(x^n) = \phi(e) = e$, so $8$, the order of $y$, divides $n$). So write $n=8k$. Then just note that since the order of $x$ is $8k$, then the order of $x^k$ is $8$, the same as the order of $y$.
A: You're almost there!
Because now, consider the group $\langle x\rangle\leq G$ generated by $x$. This group is cyclic, what do you know about its order?
Finally, use that finite cyclic groups of order $n$ have at least (in fact precisely) one cyclic subgroup of order any divisor of $n$. 
