So I'm stuck on a problem and I'd like to do more research. But I'm afraid I don't know what to even type into Google to start. The problem is this:
Let $\phi: G_1 \rightarrow G_2$ be a homomorphism of groups, $a \in G_1$, and $k \in \mathbb{Z}_{\geq 1}$. Then show that
$$\begin{align} \{ x \in G_1 \ : \ x^k = a\} &\rightarrow \{ y \in G_2 \ : \ y^k = \phi(a) \}, \\ x &\mapsto \phi(x)\end{align}$$
is well-defined.
This feels to me like the kind of question a knowledgable algebraist would look at and think, "Oh of course! This an example of the well-studied phenomenon ______." I just need a keyword so I can do further research. Does anyone know what to call a function like the one above? Or is this just some special example my professor cooked up?
Thanks in advance!
FWIW: I think the trick here is that $x$ could have finite order $n$, in which case you can write $x = x^{mn +1}$ where $m \in \mathbb{Z}$. So you have to show that $\phi(x) = \phi(x^{mn+1})$. And then as a frustrated attempted at doing that, I've written down that $\phi(x^{mn+1}) = \phi(x^{mn})\phi(x) = \phi(e_1)\phi(x) = e_2 \phi(x)$ so that indeed $\phi(x^{mn+1}) = \phi(x)$. But I'm not convinced that's right because I'm essentially just converting $x^{mn+1}$ back to $x$, rather than sending it through $\phi$ first.
Put $M := \{x \in G_1 | x^k = a \} \subseteq G_1$ and $N := \{y \in G_2 | y ^k = \phi(a) \} \subseteq G_2$.
Since $\phi: G_1 \rightarrow G_2$ is a well defined map, we certainly have $\phi(M) \subseteq G_2$.
The problem is to show that $\phi(M) \subseteq N$.
But we get this from basic properties of homomorphisms as follows.
Pick any $x \in M$. Then we have $x^k = a$. Applying $\phi$ to both sides of this equation, we get $\phi(x^k) = \phi(a)$.
Moreover, since $\phi$ is a homomorphism, we have $\phi(x^k) = \phi(x)^k$.
Combining the preceding two equations, we get $\phi(x)^k = \phi(x^k) = \phi(a)$.
But this says that $\phi(x) \in N$.
Since $x \in M$ was arbitrary, we have shown $$ \forall x \in M: \phi(x) \in N. $$ Or, in other words $$ \phi(M) \subseteq N, $$ which is what we wanted to prove.