# Let $G$ be a group and $A\subseteq G$. Showing $\{g\in G: gag^{-1}\in A \text{ for all } a\in A\}$ does not have to be a subgroup of $G$.

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 $$g^{-1}\notin H$$.

• 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. – Arturo Magidin Aug 23 at 20:17
• 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. – saulspatz Aug 23 at 20:19
• Which text are you quoting? – Shaun Aug 23 at 20:24
• @Shaun It's just the official solution to a problem from an algebra class I've taken last semester. – l2poca Aug 24 at 20:04

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 $$A = \{f \in G;~f(n) = n\quad\forall n \le 0\}.$$

• Some devices do, yes; mine didn't render two standard unicode codes earlier today. – Shaun Aug 23 at 20:20
• @k.stm: A lot of people do, and is device-dependent. – Arturo Magidin Aug 23 at 20:20
• @k.stm: At least it means you shouldn't wonder, or take offense, if someone edits your posts to switch the MathJax. – Arturo Magidin Aug 23 at 20:29
• @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. – k.stm Aug 23 at 20:35
• @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... – Arturo Magidin Aug 23 at 20:43