# Establishing a bijection between subgroups of the domain that contain the kernel and subgroups of the codomain

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.

• 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.
– jgon
Sep 11, 2020 at 23:04
• @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 $\phi^{-1}(\overline{S_1}) \neq \phi^{-1}(\overline{S_2})$.) Sep 12, 2020 at 0:43

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 $$K$$).

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.