# Inverse bijections between subgroups of $G$ containing the kernel and subgroups of $H$ contained in the image.

Claim: Let $$\varphi:G \rightarrow H$$ be a group homomorphism. Then the maps $$A \mapsto \varphi(A)$$ and $$B \mapsto \varphi^{-1}(B)$$ (where here we mean preimage) specify inverse bijections between the sets $$\{A \leq G \mid \ker(\varphi) \subseteq A\}$$ and $$\{B \leq H \mid B \subseteq \text{Im}(\varphi)\}$$.

Proof: Let $$A \leq G$$ contain the kernel, then it's evident that $$A \subseteq \varphi^{-1}(\varphi(A))$$. Let $$g \in \varphi^{-1}(\varphi(A))$$, then this means that $$\varphi(g)= \varphi(a)$$ for some $$a \in A$$. Then because $$A$$ is a subgroup $$\varphi(ga^{-1}) = \varphi(g)\varphi(a^{-1}) = 1_H$$ and so $$ga^{-1} \in \ker(\varphi) \subseteq A$$, but then because $$A \leq G$$ we have $$(ga^{-1})a = g \in A$$ and so the other inclusion holds.

The argument above isn't difficult, but I can't understand why we need to do this in the first place? If we're considering just images and preimages, doesn't it follow immediately from the definition that $$(\varphi \circ \varphi^{-1})(A) = A =(\varphi^{-1} \circ \varphi)(A)$$? What about the statement necessitates we prove it via equality of sets? Thanks in advance for the clarification.

• Here $\phi^{-1}$ is not the inverse of $\phi$, but inverse image. For example, if you compute $\phi(\phi^{-1}(A))$ [let's not use the notation $\phi\circ\phi^{-1}$] that might not be all of $A$, for all functions $\phi$ and all subsets $A$. Similarly $\phi^{-1}(\phi(A))$ can be larger than $A$.
– plop
Oct 5, 2022 at 20:58
• For example, the function $f:\{0,1\}\to\{0,1\}$ sending $f(0)=f(1)=0$, would have, for $A=\{0,1\}$, $f(f^{-1}(A))=\{0\}$ and for $B=\{0\}$ $f^{-1}(f(B))=\{0,1\}$.
– plop
Oct 5, 2022 at 21:03
• Oh of course, my fault. Don't know why I didn't realize I was treating the preimage like a function. Thanks! Oct 5, 2022 at 21:04

1. You only did half of the work: you didn't prove that forall subgroup $$B$$ in your second set, $$\varphi(\varphi^{-1}(B))=B$$.
2. You did that first half well and (with the same argument) you could have proven that more generally $$\forall X\subset G\quad\varphi^{-1}(\varphi(X))=(\ker\varphi)X$$. Similarly, for the second "half of the work", you could easily prove more generally (and here, without using that $$\varphi$$ is a morphism) that $$\forall Y\subset H\quad\varphi(\varphi^{-1}(Y))=Y\cap\operatorname{im}\varphi$$.
3. Using the two properties $$\varphi^{-1}(\varphi(X))=(\ker\varphi)X$$ and $$\varphi(\varphi^{-1}(Y))=Y\cap\operatorname{im}\varphi$$, it is easy to construct, for most morphisms $$\varphi:G\to H$$, a subgroup $$A$$ of $$G$$ and a subgroup $$B$$ of $$H$$ such that $$A\subsetneq\varphi^{-1}(\varphi(A))$$ and $$\varphi(\varphi^{-1}(H))\subsetneq H$$. Exercise: for which morphisms do such counterexamples exist?
• I understand that the preimage isn't a function so the notation I used is wrong. Would you be willing to clarify what's wrong with my understanding in the following statement? If we consider $\varphi(A)$ this is all $\varphi(x)$ such that $x \in A$, then if we take the preimage of $\varphi(A)$ we have all the elements in $G$ that $\varphi$ takes to $\varphi(A)$, is this not just $A$? That's all I would have written to exhibit the one-sided inverse here. Oct 5, 2022 at 23:14
• It is just $A$ if $\varphi$ is injective. But else it may be bigger, because there may be some $y\notin A$ having the same image as some $x\in A$. Oct 5, 2022 at 23:17
• Thanks so much! I hadn't realized I was tacitly assuming $\varphi$ was 1-1, really cleared things up! Oct 5, 2022 at 23:21
• Just to hammer the point home, to start the second part would you argue the following; $\varphi(\varphi^{-1}(B)) \subseteq B$ since the image of the preimage of $B$ is certainly contained in $B$ but may not be all of $B$ as $\varphi$ isn't necessarily surjective? Then let $b \in B$ and so on. Oct 5, 2022 at 23:53
• Yes ($\forall Y\subset H\quad\varphi(\varphi^{-1}(Y))=Y\cap\operatorname{im}\varphi$) Oct 5, 2022 at 23:55

It isn't true that $$(\varphi^{-1} \circ \varphi)(A)=A$$.
Suppose $$G = \mathbb{Z},\ A = \{0\},\ H = \{0\}$$ and $$\varphi(n) = 0\quad \forall n \in \mathbb{Z}$$
then $$(\varphi^{-1} \circ \varphi)(A) = G \neq A$$.

• Beware: not $\varphi^{-1}\circ\cdots$ (see comments above) Oct 5, 2022 at 21:08
• Beware again: $\varphi^{-1}$ alone has no meaning, it is not a function, just a notation for the inverse image. Put $\circ$ nowhere. Oct 5, 2022 at 21:16
• Thanks @AnneBauval that was clarifying. Oct 5, 2022 at 23:08