# Isomorphic groups vs. isomorphic subgroups

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

where:

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 $$H,K with $$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.

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.
• 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 $S_n$. – user615081 Jan 28 at 7:47
• @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. – Arturo Magidin Jan 28 at 7:57