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The original question that I wrote was:

A subgroup characteristic of the whole group

I was wishing that there is an argument that is simple enough to see that "clearness" since the book that I am reading (Isaacs' Algebra) recommended to accept a isomorphism as a map carrying any group theoretic property so that it also carries any canonical subgroup to another. For example, if $G \simeq H$, this isomorphism maps $[G^{n}, G]$ onto $[H^{n}, H]$ where we define $G^{1} = G, G^{2} = [G, G], G^{3} = [G^{2}, G]$, and so on. As some comments in the link suggested, it is easy to show that this is true for certain examples.

And I admit I am too pedantic to ask this question, but I do want to know how to write an argument for this fact. It is not so clear to me how to formulate the question, but after some thoughts, the following is my attept to rigorize this question.

Define $\mathscr{C} := \{(\tilde{G}, G):G \text{ is a group and } \tilde{G} \leqslant G\}$. We also consider a map $\alpha : \mathscr{C} \rightarrow \{true, false\}$. For example, $\alpha$ can ask if $\tilde{G}$ satisfies the normality condition in $G$ (i.e., $\tilde{G}^{g} = \tilde{G}$ for all $g \in G$). Let us write $\tilde{G} \alpha G$ to mean $\alpha(\tilde{G}, G) = true$.

Question. What is (minimal if possible) condition (with a proof) on $\alpha$ without referring an isomorphism so that $\tilde{G} \alpha G$ if and only if $\tilde{H} \alpha H$? (We have arbitrary groups $G, H$ with an isomorphism $\varphi:G \rightarrow H$ and $\tilde{G} \leqslant G$ with $\tilde{H} := \varphi(\tilde{G})$.)

What I wish from this question is to identify theorems "similar" to following more easily (as they can be corollaries of a good answer to this problem):

Theorem. If $G$ is a finite group with a unique Sylow $p$-subgroup $S$ for some prime $p$, then $S$ is characteristic in $G$.

Theorem. If $G$ is a finite group, then its fitting subgroup is characteristic in $G$.

Theorem. For any group $G$, the center $Z(G)$ is characteristic in $G$.

Theorem. For any group $G$, the commutator subgroup $G'$ is characteristic in $G$.

Please, let me know if this question is still ambiguous.

Moreover, I do not want to give $\alpha$ a condition referring to isomorphism since the goal that I want to achieve is to know about $\tilde{H} \leqslant H$ when we have an isomorphism $\varphi:G \rightarrow H$ and if we know much about $\tilde{G} = \varphi^{-1}(\tilde{H})$.

The following is close to what I am looking for, but again, it is disappointing that using this definition for the problem is self-explanatory.

Some of the theorems that I mentioned above need adjusted collection $\mathscr{C} := \{(\tilde{G}, G):G \text{ is a group satisfies proper axioms $A_{1}, \cdots, A_{k}$ and } \tilde{G} \leqslant G\}$.

These (not including 4 axioms of group theory) axioms $A_{j}$ can be abelian, finite, finite with unique Sylow $p$-group with a prime $p$, etc. However, I believe that an answer to more general question would give an idea to how to answer these adjusted problems as well.

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up vote 2 down vote accepted

It seems to me that this is simply the definition of a "group theoretic property."

The following is just a minor nitpick: $\{(\overline{G},G):G\text{ is a group and }\overline{G}\leqslant{G}\}$ is not a set. We can't have a set of all groups because we can't have a set of all sets, and this is clearly an extension of that (we could just take the subset of the form $(1,G)$).

Instead, we could say it in the following way. We say that $\alpha$ is a group theoretic property if and only if, whenever $\phi:G\rightarrow H$ is an isomorphism and $\overline{G}\leqslant G$, $\overline{G}$ has property $\alpha$ if and only if $\phi[\overline{G}]$ has property $\alpha$. (i.e. $\alpha$ is a logical statement about which is preserved under isomorphisms.)

So examples of $\alpha$'s could be $\overline{G}$ is the Fitting subgroup of $G$, $\overline{G}$ has order $n$, $\overline{G}$ is characteristic in $G$, $\overline{G}$ is normal in $G$, etc. Nonexamples would be things like $\overline{G}$ is made up of permutations (as even though finite groups are isomorphic to permutation groups, they are not necessarily literally groups of permutations).

I don't think you can define group theoretic properties without referring to isomorphisms. After all isomorphisms do exactly as much as we need to prove that two sets with binary operations behave the same way, and no more.

Learning to spot a group theoretic property may be a bit of an art, but the definition itself cannot get any simpler as far as I can see.

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So do you think we should just accept the idea of "relabeling" and no more? That is, the elements $a, b, c \in G$ "behave" as $ab = c$ simultaneously when $\varphi(a)\varphi(b) = \varphi(c)$ in $\varphi(G)$ where $\varphi$ is an isomorphism. – user123454321 Jan 15 '13 at 0:09
The only reason for me to be pedantic is that I somehow feel like this idea is circular. Say, I would like to know that the fitting group of a group is characteristic. But then with this idea, isn't the answer to this question too trivial and even circular? – user123454321 Jan 15 '13 at 0:11
We can intuitively think of it as a relabelling, however all of these are explicitly provable by isomorphisms. For example, $Z(G)$ is characteristic because if $ab=ba\forall b\in G$ whenever $a\in Z(G)$, then $\phi(ab)=\phi(ba)$ so $\phi(a)\phi(b)=\phi(b)\phi(a)$, and since isomorphisms are bijective $\phi(a)\in Z(H)$. Intuitively, we could have seen this simply by noting that the center of a group is defined through a holistic group theoretic statement, but this is simply interpretation - the proof is in the isomorphism. – Alexander Gruber Jan 15 '13 at 2:34
The problem that I had is that for the center, it is so simple to write down the proof, but it is not so clear that for two finite groups $G \simeq H$, the largest normal nilpotent subgroup of $G$ is mapped to that of $H$ with the same isomorphism. – user123454321 Jan 15 '13 at 4:01
However, I am very comforted by the fact that when you used the word "interpret" since if we READ $\varphi(a)$ as $a$ for all $a \in G$, we must do the same for the fitting subgroup of $G$. I think the semantic of isomorphic objects settle this. I do agree that it is more of "how they are" than "what they are proved to be." Thanks, I select your answer, as I want to move on from this thought. – user123454321 Jan 15 '13 at 4:06

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