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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.

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    $\begingroup$ Injectivity is already true on the level of subsets because the map is a surjection. Let $x$ be in one subgroup but not the other (or subset). Then $x=\phi(g)$ for some $g$ by surjectivity. Thus $g$ is contained in one preimage, but not the other. $\endgroup$
    – jgon
    Commented Sep 11, 2020 at 23:04
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    $\begingroup$ @jgon Thank you, I forgot about using surjectivity of $\phi$. That helped me show injectivity of the map $f$ that I considered. (If $\overline{S_1} \neq \overline{S_2}$, say $x \in \overline{S_1} \backslash \overline{S_2}$, then by surjectivity of $\phi$ there is $g$ such that $x = \phi(g)$. In other words, $g \in \phi^{-1}(\overline{S_1})$, but $g \notin \phi^{-1}(\overline{S_2})$. So $\phi^{-1}(\overline{S_1}) \neq \phi^{-1}(\overline{S_2})$.) $\endgroup$
    – twosigma
    Commented Sep 12, 2020 at 0:43

2 Answers 2

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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 $K$).

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Let me suggest a perhaps more direct approach. I will use a slightly different notation. Consider a surjective group morphism $f \colon G \to G'$ with kernel $\mathrm{Ker}f=K$. Introduce first the collections of subgroups:

$$\begin{align*} \mathscr{A}&\colon=\{H \leqslant G|\ K \leqslant H\}\\ \mathscr{B}&\colon=\left\{H'|\ H' \leqslant G'\right\}, \end{align*}$$ regarded as ordered sets when equipped with the natural order given by inclusion.

Recall the fact that for any $H \leqslant G$ it holds that $f[H] \leqslant G'$ (simply by virtue of $f$ being a morphism) and also in the inverse direction for any $H' \leqslant G'$ it holds that $K=f^{-1}\left[\left\{1_{G'}\right\}\right] \leqslant f^{-1}\left[H'\right] \leqslant G$ (once again since $f$ is a morphism and since trivially $\{1_{G'}\} \leqslant H'$ and taking preimages implements a map which is isotonic with respect to inclusion).

This means that you can define the following two maps: $$\begin{align*} \Phi \colon \mathscr{A} &\to \mathscr{B}\\ \Phi(H)\colon&=f[H]\\ \Psi \colon \mathscr{B} &\to \mathscr{A}\\ \Psi\left(H'\right)\colon&=f^{-1}\left[H'\right], \end{align*}$$ maps which are easily seen to be isotonic (increasing) with respect to inclusion.

It will suffice to establish the relations $\Psi \circ \Phi=\mathbf{1}_{\mathscr{A}}$ respectively $\Phi \circ \Psi=\mathbf{1}_{\mathscr{B}}$ in order to show that the two maps are mutually inverse order automorphisms and hence that $\mathscr{A} \approx \mathscr{B} \ (\mathbf{Ord})$ (by which I mean that $\mathscr{A}$ and $\mathscr{B}$ are isomorphic objects in the category $\mathbf{Ord}$ of ordered sets, fact which is known as the correspondence theorem for subgroups; there is also a version for normal subgroups as well as many other variations).

Let us proceed to establish the forementioned relations. Since $f$ is surjective, we have that $f\left[f^{-1}[Y]\right]=Y$ for any subset $Y \subseteq G'$, a general property characterising surjectivity for any map. This applies in particular to any subgroup $H' \in \mathscr{B}$ and entails the relation $\Phi \circ \Psi=\mathbf{1}_{\mathscr{B}}$.

As for the dual relation, we recall the general fact that $f^{-1}\left[f[X]\right]=KX=XK$ for any subset $X \subseteq G$ (in general, for arbitrary subsets $X, Y \subseteq G$ by $XY:=\{xy\}_{\substack{x \in X\\y \in Y}}$ I am referring to the subset product). In particular, for a subgroup $H \in \mathscr{A}$ we will have $f^{-1}\left[f[H]\right]=HK=H$ (since $K \subseteq H$ it follows that $H=H\{1_G\} \subseteq HK \subseteq HH \subseteq H$). This signifies that $\Psi \circ \Phi=\mathbf{1}_{\mathscr{A}}$.

I would like to leave you with the remark that in some instances it is possible to exhibit in a very natural manner the inverse map -- with immediate proof that it is indeed the inverse -- rather than to just argue for the bijectivity of the originally given map.

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