A theorem from the theory of groups

Question

Let $K$ be a (not necessarily normal) subgroup of the group $G$ : $K < G$

A fixed element $g\in G$ can act, from the left, on all elements of $G$, thus generating a bijection of $\,G\,$ onto itself: $\,g:\,g'\rightarrow gg'\,$.

Thereby each $g\in G$ also generates a mapping ${\hat{{L}}}_g\colon g'K \rightarrow (gg')K$, which is a bijection of the factor space $\,G/K\,$ onto itself: $$g\colon G\rightarrow G \implies \hat{L}_g\colon G/K \rightarrow G/K.$$

Let $\mathbb{L}$ be the entire set of operators $\hat{L}_g$, for all elements $g$. Then the above formula becomes: $$\mathbb{L}\colon G \rightarrow \operatorname{Aut_{set}}(G/K) ,$$

where $\operatorname{Aut_{set}}(G/K)$ denotes all bijections of $G/K$ onto itself. The subscript emphasises that $G/K$ is being mapped onto itself as a set, not as a group. (Recall that $K$ is not necessarily a normal subgroup, so $G/K$ is not required to be a group.)

Be mindful that, while each mapping ${\hat{{L}}}_g\colon g'K \rightarrow (gg')K$ is bijection, the mapping
$\,\mathbb{L}\colon G \rightarrow \operatorname{Aut_{set}}(G/K)\,$ is not even a surjection, because the surjections $\,{\hat{{L}}}_g\,$ make only a subset of the entire set $\operatorname{Aut_{set}}(G/K)$.

I need to prove the following:

Theorem. $\mathbb{L}$ is a monomorphism (i.e. $\ker\mathbb{L}=1$) if and only if $K$ contains no proper invariant subgroups of $G$.

Proving this in one direction is relatively simple, and I shall now show how to do this. Proving the inverse has turned out to be less trivial, and this is where I shall ask for your help.

Here is the one-way proof.

Let $H$ be is the set of all elements in $G$, which leave each coset of $G/K$ invariant: $$H\equiv \ker\mathbb{L}.$$

Then the following five facts will take place:

1. $H$ is a group. This is trivial.

2. $H\subset K$. This is because $H$ leaves all cosets $gK$ unmoved — including $K$.

3. $H \lhd G$. To see this, take an arbitrary $h\in H$ and then act with $\hat{L}_{g^{-1}hg}$ on some $g'K$. Instead of $\hat{L}_{g^{-1}hg}$, I write simply $g^{-1}hg$: $$(g^{-1}hg)g'K = g^{-1} h (gg'K) = g^{-1} (gg'K) = g'K ,$$ where we kept in mind that $h\in H$, so $h$ must stabilise any coset $gg'K$. The above formula demonstrates that $g^{-1}hg$ stabilises any $g'K$, so $g^{-1}hg\in H$. In other words, $g^{-1}Hg=H$.

4. $H$ is the maximal normal subgroup of $G$, contained in $K$. Suppose there exists a bigger normal subgroup $H'$ contained in $K$: $$H<H'<K,\quad H\lhd G, \quad H'\lhd G.$$ Consider a group element $x$ which is in $H'$ but not in $H$. The latter implies that $\hat{L}_x$ does not stabilise $G/K$: $$x\in H',x \notin H \implies \exists g, xgK\neq gK,$$ or, the same: $$\exists k\in K,xgk\notin gK\iff g^{-1}xgk\notin K,$$ the latter being in contradiction with the proposition that $H'\lhd G$.

5. If $G$ is isomorphic to its image in $\operatorname{Aut_{set}}(G/K)$, then $H\equiv \ker\mathbb{L}=1$ and, therefore, there exist no proper invariant subgroups of $G$, contained in $K$.

This indeed follows from the item (4) above.

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So far so good.

Now, can someone please help me to prove the inverse to the item (5)? Suppose I know that there are no proper invariant subgroups of $G$, contained in $K$. How to derive from this that $G$ is isomorphic to its image in the group $\operatorname{Aut_{set}}(G/K)$ of operators, induced by $G$ on the quotient space $G/K$?

I know that my second question will be ridiculous, but let me nevertheless ask it. When saying that $G$ is isomorphic to its image in the group $\operatorname{Aut_{set}}(G/K)$, we evidently imply that the isomorphism is implemented as follows: we choose some $g\in G$ and presume that it corresponds to $\hat{L}_g \colon g'K\rightarrow (gg')K$. Is it possible to arrange for some different isomorphism, so that $g$ will, generally, not correspond to the operator $\hat{L}_g$ generated by it? (I am asking this, because the trivial isomorphism rendered $\ker{\mathbb{L}}=1$, which may not be the case for a nontrivial isomorphism if one exists.)

Many thanks,

Michael Efroimsky

Answer 1

Let $H< K$ with $H\lhd G$. Then $G/H$ acts on $(G/H)/(K/H)$ by left multiplication. From the isomorphism theorems $(G/H)/(K/H)\approx G/K$ and the original action factors as $$\mathbb L\colon G\to G/H\to \operatorname{Aut}_{\mathbf{Set}}(G/K)$$ (because we talk about the same multiplication after all). Especially, $H\subseteq \ker\mathbb L$. Hence if $K$ contains a normal subgroup of $G$, then $\mathbb L$ is not a monomorphism.

On the other hand, let $H=\ker\mathbb L$. Then $H<K$ because we need $hK=K$ for all $h\in H$. And as kernel of a homomorphism, clearly $H\lhd G$.

Answer 2

I think I now know a simpler answer to my question.

Were $G$ not isomorphic to its image in the group $\operatorname{Aut_{set}}(G/K)$ of operators, the kernel $H\equiv \ker\mathbb{L}$ of the mapping $\mathbb{L}\colon G \rightarrow \operatorname{Aut_{set}}(G/K)$ would be different from $1$. This kernel, however, is a normal subgroup of $G$, and it also belongs to $K$. So this would contradict our assumption that there is no proper invariant subgroup of $G$, contained in $K$.

This way, the nonexistence of a proper invariant subgroup of $G$, contained in $K$, makes $\mathbb{L}$ monomorphism.

Once again, many thanks to everyone,

Michael Efroimsky