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.

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

  • $\begingroup$ but surely if you take a finite group $G$ of order at least 3 and $K={1}$, $K$ contains no proper normal subgroups but $G$ is not isomorphic to $\text{Aut}_\text{set}(G)$ (which is just $S_{|G|}$ and hence has order $|G|! > |G|$), contradicting your theorem. Are there extra assumptions on $K$ and $G$ that you've left out, or am I missing something? $\endgroup$
    – Astrid
    Nov 28, 2014 at 21:06
  • $\begingroup$ Thank you Astrid. Does the further comment by Hagen von Eitzen on monomorphism vs isomorphism resolve this? $\endgroup$ Nov 28, 2014 at 22:05
  • $\begingroup$ Yes, that explains it! Monomorphism makes a lot more sense than isomorphism in that statement. $\endgroup$
    – Astrid
    Nov 28, 2014 at 23:32
  • $\begingroup$ Thank you Astrid! I learned a lot today. $\endgroup$ Nov 29, 2014 at 2:02
  • $\begingroup$ I reformated your post. You shouldn't impose your spacing, LaTeX is already taking care of it and does a (much much) better job ! Just focus on the ideas, the form is being handled for you. $\endgroup$
    – Pece
    Dec 2, 2014 at 6:45

2 Answers 2


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 $H\lhd G$.

  • $\begingroup$ Many thanks! I should cogitate on this. Can you kindly have a look at my second question? $\endgroup$ Nov 28, 2014 at 21:00
  • $\begingroup$ @Michael_1812 There seems to be some misunderstanding here (in fact, an abuse of the term "isomorphism" in the problem statement) $\mathbb L$ is a monomorphism. i.e. an isomorphism with its image only, not with the full $ \operatorname{Aut}_{\mathbf{Set}}(G/K)$ (or even $\operatorname{Aut}_{\mathbf{Grp}}(G/K)$ if $K\lhd G$). But it is ingeneral not an isomorphism. For example if $K$ is the trivial subgroup of a finite group $G$ of $n$ elements, then $\operatorname{Aut}_{\mathbf{Set}}(G/K)$ has $n!$ elements, whihc is $>n$ for $n\ge 3$. $\endgroup$ Nov 28, 2014 at 21:42
  • $\begingroup$ Yes, you are right. Does this observation resolve the issue pointed above by Astrid? $\endgroup$ Nov 28, 2014 at 22:18
  • $\begingroup$ May I return to my second question. $G$ is monomorphic to a subgroup of $\operatorname{Aut_{set}}(G/K)$. The subgroup is defined like this: choose some $g\in G$ and set it to correspond to $\hat{L}_g \colon g'K\rightarrow (gg')K$. Is it possible to build a different monomorphism, which maps $G$ onto the same subset of $\operatorname{Aut_{set}}(G/K)$, though by a different rule (so $g$ will, generally, not correspond to the operator $\hat{L}_g$ defined as above)? While the map $\hat{L}_g$ satisfies $\ker{\mathbb{L}}=1$, this may not be the case for a nontrivial monomorphism if it exists. $\endgroup$ Dec 3, 2014 at 0:59
  • $\begingroup$ @Michael_1812 If $\phi,\psi\colon G\to H$ are group monomorphism with the same image, then they differ only by an automorphism of $G$, i.e., $\psi=\phi\circ\alpha$ for some $\alpha\in\operatorname{Aut}_{\mathbf{Grp}}(G)$. Anyway, by the very condition that the image should be the same, any property holding for all $\phi(g)$ also holds for all $\psi(g)$. $\endgroup$ Dec 3, 2014 at 11:09

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


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