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If $f \in \mathrm{Aut}(G)$, in particular $f$ is bijective. So, $|f(H)|=|H|$. Since $f$ is an isomorphism, then $f(H) \cong H$?

user5826
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    It implies $f(H)\cong H$, but not by cardinality arguments (which you cannot use when $H$ is infinite, which is a possibility your hypotheses do not rule out). However, note that $f^{-1}$ is also an automorphism, so $f^{-1}(H)\subseteq H$, and applying $f$ to both sides you get $f(H)=H$. Then you can conclude that $f|_H$ is a bijection and hence an isomorphism. – Arturo Magidin Jun 30 '19 at 01:16
  • @ArturoMagidin Ok I'm a bit more confused. If $H$ is characteristic, then $f(H)=H$ for all $f \in \mathrm{Aut}(G)$? Why it defined as $f(H) \subset H$? – user5826 Jun 30 '19 at 01:26
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    Because both definitions are equivalent, but you want to have your definition be the weakest you can get away with, as it makes it easier to check. It is much easier to check that for every $f$ you have $f(H)\subseteq H$ than you check that for every $f$ you have $f(H)=H$. Since the former implies the latter, then define the condition as the former, knowing you will get the full condition as a consequence. – Arturo Magidin Jun 30 '19 at 01:28
  • @ArturoMagidin Ahh I see now. Got it. – user5826 Jun 30 '19 at 01:29
  • It's the same as for normal subgroups, where requiring $xHx^{-1}\subset H$ is enough (it is good to note, however, that it's not true that "for all $f\in Aut(G)$, ($f(H)\subset H\implies f(H)=H$)" : you have to have inclusion for every automorphism in order to deduce equality for every automorphism) – Maxime Ramzi Jun 30 '19 at 10:19
  • @Max Why is that statement not true? As Arturo writes, $f^{-1} \in Aut(G)$ and so $f^{-1}(H) \subset H$ also. But then applying $f$ we have $H \subset f(H)$. So, $f(H)=H$. – user5826 Jun 30 '19 at 15:54
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    There is a difference between "(for all $f\in Aut(G), f(H)\subset H) \implies (\forall f \in Aut(G), f(H)=H)$" , which is the statement Arturo proved and the one you proved again; and "for all $f\in Aut(G), (f(H)\subset H\implies f(H)=H)$" - of course this second statement was not the object of your original question, I was just pointing it out as a side note – Maxime Ramzi Jun 30 '19 at 15:56
  • @Max On second thought, I don't really understand why the second statement is not true. It seems like it's just a less precise rewording of the first statement. If for every automorphism $f$ of $G$, we have $f(H) \subset H$, can we show $f(H)=H$? It seems like its the same as the first statement. Can you clarify a bit? – user5826 Jun 30 '19 at 16:04
  • Look at the parentheses. The second statement says "give me an automorphism. Now if for this specific automorphism $f$, we have $f(H)\subset H$, then for this specific $f$ we have $f(H)=H$." Arturo's argument can't apply here because of course we haven't assumed that $f^{-1}(H)\subset H$ ($f^{-1}$ is a different automorphism than $f$ in general). Look for instance at $\bigoplus_{\mathbb Z} \mathbb Z$ with automorphism given by a shift on the index, that is, $e_n$ gets sent to $e_{n+1}$. Then the people in nonnegative degree stay there, but $e_0$ for instance is not in their image – Maxime Ramzi Jun 30 '19 at 16:11
  • (where $e_n$ is the generator $1$ of $\mathbb Z$ in position $n$) – Maxime Ramzi Jun 30 '19 at 16:12
  • Note: for any automorphism $f$ and any subgroup $H$, you always have $H\cong f(H)$, since $f$ is a bijective group homomorphism. Presumably, you mean equality, not isomorphism. – Arturo Magidin Jun 30 '19 at 19:33

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Lest this keep being in comments, let me add some words here.

First: if for every $f\in\mathrm{Aut}(G)$ we have $f(H)\subseteq H$, then we can conclude that for every $f\in\mathrm{Aut}(H)$ we have $f(H)=H$; however, the cardinality argument is only valid when $H$ is finite.

To get the conclusion in general, let $f\in\mathrm{Aut}(G)$. By assumption we already know that $f(H)\subseteq H$. To prove equality, consider $f^{-1}\in\mathrm{Aut}(G)$; by assumption we have $f^{-1}(H)\subseteq H$, since the condition applies to all automorphisms. Applying $f$ to both sides we have $f(f^{-1}(H))\subseteq f(H)$. Now, since $f$ is an automorphism, $ff^{-1}=\mathrm{id}_G$, so we conclude $H\subseteq f(H)$, giving the desired equality.

(From this, since $f$ is bijective, we also get $H\cong f(H)$, but of course this holds for any automorphism, whether $f(H)\subseteq H$ or not).

Now, as Max notes in comments, one has to be careful: it is not the same to say that “(for all $f\in\mathrm{Aut}(H)$, $f(H)\subseteq H$) implies (for all $f\in\mathrm{Aut}(H)$, $f(H)=H$)” than to say $$\forall f\in\mathrm{Aut}(G)\Bigl( f(H)\subseteq H \implies f(H)=H\Bigr).$$ Why? This is weaker, since it says nothing about what happens to $H$ under automorphisms that do not send $H$ into $H$. For example, in the Klein $4$-group, no nontrivial proper subgroup is characteristic: none of them satisfy that $f(H)\subseteq H$ for all automorphisms. However, all nontrivial proper subgroups satisfy that if $f\in\mathrm{Aut}(G)$ happens to send $H$ into $H$, then it maps $H$ to itself identically. And more generally, the implication holds in any finite group for any subgroup, characteristic or not.

Now, the second question: if the condition $$\forall f\in\mathrm{Aut}(G)\Bigl( f(H)\subseteq H\Bigr)$$ is equivalent to asking that $f(H)=H$, why not use the latter condition instead?

Two reasons: one is simplicity. It is better to have the weaker condition to check than the stronger one, because it will make it easier to use. It is easier to verify that $f(H)\subseteq H$ for all $f\in\mathrm{Aut}(G)$, than to verify that $f(H)=H$.

The second reason is that it fits better into the general framework of normal, characteristic, and fully invariant subgroups: let $G$ be a group, and let $H$ be a subgroup.

  1. $H$ is normal in $G$ if and only if for every $f\in\mathrm{Inn}(G)$, $f(H)\subseteq H$.
  2. $H$ is characteristic in $G$ if and only if for every $f\in\mathrm{Aut}(G)$, $f(H)\subseteq H$.
  3. $H$ is fully invariant in $G$ if and only if for every $f\in\mathrm{End}(H)$, $f(H)\subseteq H$.

While the first two can be defined by requiring equality rather than inclusion, the third cannot: it must be defined using inclusion only.

(I did a brief discussion of normal, characteristic, fully invariant, and verbal subgroups in this answer.)

Arturo Magidin
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