Let $\phi: G \to \overline{G}$ be a surjective homomorphism with kernel $K$. I am wondering if there is a bijection from the collection of subgroups of $G$ that contain $K$, $\{ S: S \leq G, S \supseteq K \}$, and the collection of subgroups of $\overline{G}$, $\{\overline{S}: \overline{S} \leq \overline{G} \}$.
My attempt:
Let $A := \{ S: S \leq G, S \supseteq K \}, B := \{\overline{S}: \overline{S} \leq \overline{G} \}$. I considered the function $f: B \to A$ that sends $\overline{S}$ to $\phi^{-1}(\overline{S})$, the preimage of $\overline{S}$ under $\phi$. We need to check three things: that this function is well-defined, that it is surjective, and that it is injective.
[TL;DR: I think I have showed well-defined-ness and surjectivity, but I'm not sure about injectivity.]
Well-defined: Given a subgroup $\overline{S}$ of $\overline{G}$, we need to show that its preimage (under $\phi$), call it $S$, is a subgroup of $G$ that contains $K$. This would show that $f$ is indeed well-defined. $S$ contains $K$ because given $g \in K$, we have $\phi(g) = \overline{e} \in \overline{S}$. Since $K$ is nonempty, so is $S$, and so to show that $S$ is a subgroup, all that remains is to show closure under the operation and closure under inverses. Thus, let $x, y \in S$. Then $\phi(xy) = \phi(x)\phi(y) \in \overline{S}$, which shows closure under the operation. Finally, $\phi(x^{-1}) = \phi(x)^{-1} \in \overline{S}$, which shows closure under inverses.
Surjective: Given $S \leq G, S \supseteq K$, we want to show that $S$ is the preimage of some subgroup $\overline{S}$ of $\overline{G}$. I claim that the image of $S$, $\phi(S)$, satisfies this. That is, I claim that $\phi(S)$ is a subgroup of $\overline{G}$ and $\phi^{-1}(\phi(S)) = S$. First, we prove that $\phi(S)$ is a subgroup of $\overline{G}$. Well, $\phi(S)$ is nonempty because it contains $\overline{e} = \phi(e)$. It is closed under the operation because given $s, t \in S$, we have $\phi(s)\phi(t) = \phi(st) \in \phi(S)$. It is closed under inverses because $\phi(s)^{-1} = \phi(s^{-1}) \in \phi(S)$. Next, we show that $(\phi^{-1}(\phi(S)) = S$. The containment $S \subseteq (\phi^{-1}(\phi(S))$ is a standard fact about images/preimages. As for $(\phi^{-1}(\phi(S)) \subseteq S$, suppose $x$ is in the LHS, so that $\phi(x) \in \phi(S)$, so that $\phi(x) = \phi(s)$ for some $s \in S$. So $\overline{e} = \phi(x)\phi(s)^{-1} = \phi(xs^{-1})$, hence $xs^{-1} \in K$. Since $K \subseteq S$ by assumption, this means $xs^{-1} \in S$, so $x \in S$.
Injective: This is where I am stuck. I think I should show that no subgroup $S$ of $G$ that contains $K$ can be the preimage of two different subgroups of $\overline{G}$, which I've tried to do but am not sure how.
Best Answer
Yes, this is true and part of the isomorphism theorems. Show that every subgroup containing $K$ is the preimage of its image (this property characterizes the subgroups containing $K$).