Does $k$ generate $K$? Suppose that $K$ is a finite cyclic group, let $G$ be any group, and suppose that $\phi\colon K\to G$ is a homomorphism. I'm trying to determine whether or not, given $k\in K$ such that $\phi(k)$ generates $\phi(K)$, can I conclude that $k$ generates $K$?
I know that for all $j\in K$, there exists $n\in\mathbb Z$ such that
\begin{align*}
\phi(j) & = \phi(k)^n \\
\phi(j) & = \phi(k^n)
\end{align*}
but as I don't know that $\phi$ is injective, I get stuck here.
 A: The issue is precisely that $\varphi$ is not injective. What if $\varphi$ sends everything to the identity? Clearly the identity of $G$ generates $\varphi(K)$, but not any element of $K$ generates $K$. 
I thought at first that the map had to be injective for the statement to hold, but actually no : the map $\mathbb Z / 4 \mathbb Z \to \mathbb Z / 2 \mathbb Z$ sends $3$ to $1$ and $3 = -1 \pmod 4$ generates $\mathbb Z / 4 \mathbb Z$. But injectivity of course ensures that the result holds. My point is that it is possible that your statement holds even in the non-injective case. The key is that if $k \in \mathbb Z$, $(k,m) = 1$ implies $(k,n) = 1$ as long as $(k,n/m) = 1$, which can be satisfied by numerous pairs of integers ; the greatest common divisor condition is the key to generating $\mathbb Z / n \mathbb Z$. To sum things up, this argument proves : 
$k \in \mathbb Z / n \mathbb Z \overset{def}= K$ generates $K$ if and only if $\varphi(k)$ generates $\varphi(K)$ and $(k, |K|/|\varphi(K)|) = 1$. 
Hope that helps,
A: You have correctly shown that this is true if $\phi$ is injective. It is false if $\phi$ is not injective.
For a trivial counterexample, let $$\phi:K\to G\\k\mapsto e$$
be the trivial homomorphism. Then, for example, $\phi(e)$ generates $\phi(K)$, but clearly $e$ doesn't generate $K$.
