Let's consider the following diagram: $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\las}[1]{\kern-1.5ex\xleftarrow{\ \ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} I_n & & \color{red}{I_m} & & \color{red}{I_m} & & I_n \\ \da{f} & & \color{red}{\da{\hat f}} & & \color{red}{\da{\bar f}} & & \da{f} \\ G & \las{\hat\epsilon} & \color{red}{H} & \color{red}{\ras{\psi}} & \color{red}{K} & \ras{\bar\epsilon} & G \\ \da{\theta} & & \color{red}{\da{\hat\theta}} & & \color{red}{\da{\bar\theta}} & & \da{\theta} \\ S_G & \las{\hat\iota} & \color{red}{S_H} & \color{red}{\ras{\varphi^{(\psi)}}} & \color{red}{S_K} & \ras{\bar\iota} & S_G \\ \da{\varphi^{(f)}} & & \color{red}{\da{\varphi^{(\hat f)}}} & & \color{red}{\da{\varphi^{(\bar f)}}} & & \da{\varphi^{(f)}} \\ S_n & & \color{red}{S_m} & & \color{red}{S_m} & & S_n \\ \end{array} $$


  1. $m,n$ are positive integers
  2. $I_x:=\{1,\dots,x\}$, for $x=m,n$
  3. $\hat f$ is a bijection
  4. $G$ is a finite group of order $n$, and $H,K<G$ with $H\ne K, H \cong K$
  5. $\psi$ is an isomorphism
  6. $\bar f:=\psi \hat f$
  7. $S_X:=\operatorname{Sym}(X)$, for $X=G,H,K$
  8. $\theta,\hat\theta,\bar\theta$ are Cayley embeddings
  9. $S_x$ is the symmetric group of degree $x$, for $x=m,n$
  10. $\hat\epsilon,\hat\iota,\bar\epsilon,\bar\iota$ are embeddings such that: $$\hat\iota\hat\theta=\theta\hat\epsilon, \quad \bar\iota\bar\theta=\theta\bar\epsilon \tag 0$$
  11. given two sets $A,B$ and a bijection $\alpha\colon B \rightarrow A$, the map $\varphi^{(\alpha)}\colon S_A \rightarrow S_B$ is the isomorphism defined by $\sigma \mapsto (b \mapsto (\alpha^{-1}\sigma\alpha)(b))$.

If we single out the red-coloured part of the diagram, and interpret $H$ and $K$ as independent entities, then this answer has already shown that:

$$\varphi^{(\hat f)}\hat\theta \hat f=\varphi^{(\bar f)}\bar\theta \bar f \tag 1$$

Namely: two isomorphic (abstract) groups of order $m$ can embed into one same subgroup of $S_m$. In this sense, "they need not be distinguished from the standpoint of group theory" (see this other answer).

Now, I would like to see what differences get in if $H$ and $K$ are no longer "independent entities", but rather subgroups of the parent group $G$ (whole diagram). In particular, can $H$ and $K$ embed into one same subgroup of $S_n$ via $S_G$? Equivalently: Does there exist a bijection $f\colon I_n \to G$ such that: $$\varphi^{(f)}\theta\hat\epsilon \hat f = \varphi^{(f)}\theta\bar\epsilon \bar f \tag 2$$?

Hoping to have posed well this kind of question.


No, you won't be able to do this in general.

For an example, take $G$ to be the dihedral group of order $8$, $G=\langle r,s\mid r^4=s^2=1,\ sr=r^3s\rangle$, let $H=\langle r^2\rangle$ and $K=\langle s\rangle$. Both $H$ and $K$ are cyclic of order $2$, and thus abstractly isomorphic to one another. Note that not only are they not conjugate in $G$, but also they are not conjugate in the holomorph of $G$, $G\rtimes\mathrm{Aut}(G)$, since $H$ is the center of $G$ and so always maps to itself under an automorphism.

In particular, the image of $\varphi^{(f)}\theta\hat{\epsilon}\hat{f}$ (that is, of $H$) will necessarily lie in the center $\varphi{(f)}(G)$. But the image of $\varphi^{(f)}\theta\bar{\epsilon}\bar{f}$ (which is the image of $K$) is not central in $\varphi^{(f)}(G)$, and so they cannot be equal.

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  • $\begingroup$ I'm wondering whether this is a general feature of (distinct) isomorphic subgroups of a group $G$ of order $n$, namely not to be embeddable, via $S_G$, into one same subgroup of $S_n$. $\endgroup$ – user615081 Jan 28 at 7:47
  • 1
    $\begingroup$ @Luca: If $H$ and $K$ are conjugate, then you can “twist” one of the embeddings of $G$ into $S_G$ so that they end up mapped to the same elements of $S_G$ (via the conjugation that corresponds to the conjugation in $G$). And if $H$ and $K$ are just isomorphic, then there is an overgroup of $G$ in which $H$ and $K$ are conjugate in a way that matches your initial isomorphism (it’s called an HNN extension). The problem here is that you are trying to keep $G$ fixed while also making $H$ and $K$ coincide, and that’s going to be utterly impossible in most cases. $\endgroup$ – Arturo Magidin Jan 28 at 7:57

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