Let $\phi : G\twoheadrightarrow H$. Suppose $W, K$ are conjugate in $H$. Show $\phi^{-1}(W), \phi^{-1}(K)$ are conjugate in $G$.

Question

In trying to answer this question on MSE, I got stuck. This taunts me because I think I should be able to do it.

The Question:

Let $\phi : G\twoheadrightarrow H$ be an epimorphism of groups. Suppose $W$ and $K$ are conjugate in $H$. Show that $\phi^{-1}(W)$ and $\phi^{-1}(K)$ are conjugate in $G$.

My Attempt:

Following @DerekHolt's comment . . .

It is true and straightforward $g\phi^{-1}(W)g^{\color{red}{-1}} = \phi^{-1}(K)$. Just do the normal thing and show that any element of the LHS is in the RHS and vice versa.

So here goes . . .

Let $h\in g\phi^{-1}(W)g^{-1}$ for some $g\in G$. Then there exists a $w\in W$ for which $h=g\phi^{-1}(w)g^{-1}$. But since $gWg^{-1}=K$, there exists a $k\in K$ such that . . . I don't know.

It's not immediately obvious what I should do, even given the hint. I think I did something simple wrong already.

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NB: It's 00:41 here . . . now. That's my excuse.

Answer 1

Small note: in your attempted proof you have $g \in G$ and $W \subseteq H$, so $gWg^{-1}$ isn't (a priori) a well-defined thing. Perhaps trying to work with this object made things seem more confusing than they are.

Let's make sure to use both of the assumptions: that $\phi$ is surjective (equivalent to epimorphic in the category of groups), and that $W$ and $K$ are conjugate in $H$. So, let $hWh^{-1} = K$ for some $h \in H$ and let $g \in G$ such that $\phi(g) = h$. We claim that $g\phi^{-1}(W)g^{-1} = \phi^{-1}(K)$. First, let $x \in g\phi^{-1}(W)g^{-1}$ be arbitrary, and write $x = g a g^{-1}$ for some $a \in \phi^{-1}(W)$. Then $\phi(a) \in W$, so $$\phi(x) = \phi(g) \phi(a) \phi(g^{-1}) = h \phi(a) h^{-1} \in h W h^{-1} = K.$$ In other words, $x \in \phi^{-1}(K)$, so (since $x$ was arbitrary) $g \phi^{-1}(W) g^{-1} \subseteq \phi^{-1}(K)$. Conversely, let $y \in \phi^{-1}(K)$ be arbitrary. Then $$\phi(g^{-1}yg) = h^{-1} \phi(y) h \in h^{-1} K h = h^{-1} h W h^{-1} h = W,$$ so $g^{-1} y g \in \phi^{-1}(W)$. Thus, $y \in g\phi^{-1}(W)g^{-1}$. Since $y$ was arbitrary, $\phi^{-1}(K) \subseteq g\phi^{-1}(W)g^{-1}$.

Answer 2

Choose $x \in \phi^{-1}(W).$ Then $\phi(x) \in W$, so $\exists h \in H \text{ such that } h\phi(x)h^{-1} \in K.$ Because $\phi$ is an epimorphism, $\exists g \in G \text{ with } \phi(g)=h$, so $\phi(gxg^{-1})=h\phi(x)h^{-1} \in K$ and $gxg^{-1} \in \phi^{-1}(K)$. Thus, $\phi^{-1}(K) \subseteq g^{-1}\phi^{-1}(W)g$. Similarly, $\phi^{-1}(W)$ is contained in a conjugate of $\phi^{-1}(K)$ so equality follows.