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Let $G$ be a group and $A\subseteq G$. Consider $H=\{g\in G: gag^{-1}\in A \text{ for all } a\in A\}$. I want to show that that $H$ does not have to be a subgroup of $G$. (Note that for a finite $G$ it always is a subgroup because it suffices to prove closure, which is true in this case).

The material I am using gives the following counterexample, I see why it is a counterexample but I would like to understand intuitively how you can arrive at this result.

Let $G$ be the set of all permutations over $\mathbb{Z}$ and define $S_f=\{n\in\mathbb{Z}:f(n)\neq n\}$ for $f\in G$. Set $A=\{f\in G: S_f\subseteq \mathbb{N}_{>0}\}$. Consider $g\in G$ with $g(n)=n+1$ for all $n$. It's easy to check that $g\in H$. To arrive at a counterexample, we now show that $g^{-1}\notin H$. Let $a\in G$ with $a(1)=2$, $a(2)=1$ and $a(n)=n$ for $n\neq 1,2$, so that $a\in A$. Then $g^{-1}ag(0)=g^{-1}a(1)=g^{-1}(2)=1$. This shows that $0\in S_{g^{-1}ag}$ and thus $g^{-1}ag\notin A$ so that $g^{-1}\notin H$.

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    $\begingroup$ You may find the example in this answer easier to grasp. Basically, the element in question "shrinks" $A$, which is fine, but the inverse tries to "enlarge" $A$ and so you may end up with something that is too big. $\endgroup$ – Arturo Magidin Aug 23 at 20:17
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    $\begingroup$ This example is much like the fact that integrating a polynomial and then differentiating gets you back where you started, but differentiating and then integrating doesn't. $\endgroup$ – saulspatz Aug 23 at 20:19
  • $\begingroup$ Which text are you quoting? $\endgroup$ – Shaun Aug 23 at 20:24
  • $\begingroup$ @Shaun It's just the official solution to a problem from an algebra class I've taken last semester. $\endgroup$ – l2poca Aug 24 at 20:04
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As you noted, $H$ is always a closed unter multiplication, thus $H \subseteq G$ is a submonoid. The only way for $H$ to fail to be a subgroup is for it to fail to contain an inverse for one of its elements.

Thus, to find a counterexample we have to ensure that no argument of the form $$gAg^{-1} \subseteq A \implies g^{-1}Ag \subseteq A$$ holds. As you already said, this argument is valid if $A$ is finite, so $A$ cannot be finite.

Think of the elements in $G$ as moves, of $A$ as a property and of elements of $A$ as moves with that property. So we need a property of moves that is preserved when first moving in one way beforehand and then moving in the opposite way afterwards, but that is not preserved the other way around - when first moving in the opposite way beforehand and then moving in the original way afterwards.

The simplest infinite set would be $\Bbb Z$, the simplest moves on $\Bbb Z$ would be jumps to the left or to the right, but more generally any permutations of $\Bbb Z$. And the simplest “ways” to move would be, well, left and right. The simplest property of a move would be fixing stuff.

So the question becomes:

Which moves on $\Bbb Z$ fix some stuff, even when first going left beforehand, then going right afterwards, but not when first going right beforehand, then going left afterwards?

Well, if a move fixes some left half of $\Bbb Z$, it certainly also does so when moving even further left before and moving back right after – but when I move right first, there’s no guarantee anymore!

So, that’s how you could come up with the counterexample . . .

Note that the way $A$ is defined in the text is a bit convoluted. A simpler definition might be $$A = \{f \in G;~f(n) = n\quad\forall n \le 0\}.$$

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    $\begingroup$ Some devices do, yes; mine didn't render two standard unicode codes earlier today. $\endgroup$ – Shaun Aug 23 at 20:20
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    $\begingroup$ @k.stm: A lot of people do, and is device-dependent. $\endgroup$ – Arturo Magidin Aug 23 at 20:20
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    $\begingroup$ @k.stm: At least it means you shouldn't wonder, or take offense, if someone edits your posts to switch the MathJax. $\endgroup$ – Arturo Magidin Aug 23 at 20:29
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    $\begingroup$ @ArturoMagidin No, of course not. The edit is fine, of course. I was serious. I think I ought to switch to latex – at least after having answered. $\endgroup$ – k.stm Aug 23 at 20:35
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    $\begingroup$ @k.stm: Good to know; the comment seemed to me a bit accusatory, but then it's really hard to convey tone in text, and "Do you have a problem [with]..." has been used in too many B-movies... $\endgroup$ – Arturo Magidin Aug 23 at 20:43

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