Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We say a subgroup $H \leqslant G$ is characteristic in $G$ if for every $\varphi \in \text{Aut}(G)$, we have $\varphi(H) = H$.

Now, suppose that we have a unique subgroup $S \leqslant G$ that has a certain property (e.g. the center of $G$). Then is it just obvious that $S$ is characteristic in $G$?

Thinking of any isomorphism as "relabeling", this is clear, but I don't know if I should try to sit down to formulate this problem correctly and write the proof of it.

I asked another question related after this: Mapping between subgroups by an isomorphism.

share|cite|improve this question
To answer this question, you'll have to make precise "unique subgroup that has a certain property". – Chris Eagle Jan 14 '13 at 15:09
Formulating the problem is a part of this question. That's why I gave an example. – user123454321 Jan 14 '13 at 15:10
$S$ also can be the commutator subgroup and in finite case, the fitting subgroup. – user123454321 Jan 14 '13 at 15:11
$H$ being characteristic is also clear if there is "only one candidate", e.g. if it is the only subgroup of order $42$ or the only normal subgroup with $G/H$ isomorphic to he baby monster group. Also if it consists of (or is generated by) all elements with a property not involving parameters. – Hagen von Eitzen Jan 14 '13 at 15:13
@HagenvonEitzen: I agree with you. So maybe I shouldn't even bother to think about it and move on? – user123454321 Jan 14 '13 at 15:14

If $i:G_1\to G_2$ is a group isomorphism, and $S \le G_1$ has some group-theoretic property $P$ in $G_1$, then also $i(S)$ has this group-theoretic property in $G_2$.

This is true whenever $P$ is some property expressible in the language of groups, in any logic (first order, second order, infinitary, etc.). This is easy to prove for any specific property (such as "is the center"), but if you want to talk about all properties you first need to specify what a "property" is.

Depending on how much logic you know, you may want to define property as "first order" or "second order" property: This is done by induction: Every equality between terms is a formula, and also "$t \in S$" is a formula whenever $t$ is a term. Then, formulas are closed under negation, conjunction, implication, quantification...

Once you have proved the above, you can apply it to case $G_1=G_2$.

(As a corollary: Every characteristic subgroup is a normal subgroup.)

share|cite|improve this answer
The corollary you mentioned can be seen by taking any inner automorphism, and it is somewhat not relevant to your answer, if I understand what you are saying not far off from your intention. – user123454321 Jan 14 '13 at 15:19
Then the main question is, how would you formulate this problem and prove it? For me, this looks like something that is very "clear" but once I try to write it down, it will take much time. – user123454321 Jan 14 '13 at 15:23
@GilYoungCheong: For first order properties, the main idea is that isomorphic models are elementary equivalent. – Seirios Jan 14 '13 at 18:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.