no, that's different, what this means is that for example, the subgroup $\{\mathrm{id},(12)(34),(13)(24),(14)(23)\}$ is a normal subgroup of $S_4$, whereas $\{\mathrm{id},(12),(34),(12)(34)\}$ is not a normal subgroup of $S_4$, althought both are isomorphic to $\Bbb Z/2\times \Bbb Z/2\Bbb Z$.
What it means is that the statement "$H$ is a normal subgroup of $G$" depends on how you embed $H$ into $G$ and not just on the abstract groups $H$ and $G$

so... he basically avoid talking about any derivatives by computing $f(x+\delta)$ and then get coefficients of this expansion, which happens to be the partial derivatives
I am not convinced that this works in general...

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💥(Semigroup)={categories,...,...,...}

I'm a bit unsure as of how to do the induction step here.

https://math.stackexchange.com/questions/996641/show-functionally-completeness-property-for-propositional-logic

So, given a group $O$ where $\{o\in O\}\bigcap\{o^{-1}\in O\}=I_O$, and $O$ has the traits:
$$\begin{align}oo=o^{-1}\\ o^{-1}o^{-1}=o\\ o_{1}o_{2}=o_{3} : o_{1}\neq o_{2}\neq o_{3}\\ o_{1}^{-1}o_{2}^{-1}=o_{3}^{-1} : o_{1}^{-1}\neq o_{2}^{-1}\neq o_{3}^{-1}\\oo^{-1}=I_o\end{align}$$
Is it possible to define $o_{1}o_{2]^{-1} such that the entire group remains abelian?