I was trying to give the wording "isomorphic groups have the same structure" a precise interpretation, at least for the finite order case. A positive answer to my final question would reach the goal, as it would translate a "$\cong$" between groups into a "$=$" between sets.
For $n$ positive integer, let be:
- $I_n:=\{1,\dots,n\}$;
- $G$, $\overline G$ groups of order $n$;
- $\psi\colon G \rightarrow \overline G$ isomorphism;
- $f$, $\bar f$ bijections;
- $\theta$, $\bar \theta$ embeddings;
- in general, $\varphi^{(\alpha)}$ the isomorphism between symmetric groups on sets of the same cardinality, defined by $\sigma \mapsto (g \mapsto (\alpha\sigma\alpha^{-1})(g))$, where $\alpha$ is a bijection between the sets;
- $S_n$ the symmetric group of degree $n$.
Visually:
$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
I_n & & I_n \\
\da{f} & & \da{\bar f} \\
G & \ras{\psi} & \overline G \\
\da{\theta} & & \da{\bar \theta} \\
\operatorname{Sym}(G) & \ras{\varphi^{(\psi)}} & \operatorname{Sym}(\overline G) \\
\da{\varphi^{(f)}} & & \da{\varphi^{(\bar f)}} \\
S_n & & S_n \\
\end{array}
$$
Is it $(\varphi^{(f)}\circ\theta)(G)=(\varphi^{(\bar f)}\circ\bar \theta)(\overline G)$?
Best Answer
No, those things need not be equal "for trivial reasons": you can take $G = G'$, and $\theta = \theta'$, and $\psi$ to be the identity (although it doesn't even factor into the equation you're asking about), but take $f \neq \bar f$. Then trivially $\theta(G) = \theta'(G')$, but the mappings $f, f'$ give different isomorphisms $\phi^{(f)}$ and $\phi^{(f')}$, so that you get different images in $S_n$. Concretely, you could take $G = C_4$ with generator $g$, and let $\theta$ be the Cayley embedding, and let $f(k) = g^k$, while $f'(1) = g, f'(2) = g^3, f'(3) = g^2, f'(4) = e$, so that $(1, 2, 3, 4)$ is in $(\phi^{(f)} \circ \theta)(G)$ but not in $(\phi^{(f')} \circ \theta')(G')$.
The problem with this is not that group isomorphism is the wrong notion, it's that you're asking the wrong question about it. It would work better with the following changes: suppose that $f, f'$ have the additional condition that $\psi \circ f = f'$, and that $\theta, \theta'$ are the Cayley embeddings for $G, G'$, then do the two mappings $I_n \to S_n$ given by $$ \phi^{(f)} \circ \theta \circ f \qquad \text{and} \qquad \phi^{(f')} \circ \theta' \circ f' $$ agree? And indeed the answer is that they do. For take $i, k \in I_n$, then $$ \begin{align*} \phi^{(f)}(\theta(f(i)))(k) &= f^{-1}(\theta(f(i))(f(k)))\\ &= f^{-1}(f(i)f(k))\\ &= f^{-1}\psi^{-1}\psi(f(i)f(k))\\ &= (f^{-1}\psi^{-1})(\psi(f(i))\psi(f(k)))\\ &= f'^{-1}(f'(i)f'(k))\\ &= f'^{-1}(\theta'(f'(i))(f'(k)))\\ &= \phi^{(f')}(\theta'(f'(i)))(k). \end{align*} $$