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I am given a group $M$ with identity element $1$, and $A$ $\leq$ $M$. I am then given a set $G = \{m \in M \mid mam^{-1} \in A, \ \forall a \in A\}$ and asked to prove that $G \leq M$.

I am confused about both the closure and identity.

Attempt for proving closure:

Let $g$, $h$ be two elements in $G$ such that $gag^{-1} \in A$ and $hah^{-1}$ $\in$ $A$. Then I have to show that $(gh)a(gh)^{-1}$ $\in$ $A$ for the same $a \in A$, right? Here is what I tried:

Take $(gh)a(gh)^{-1} \implies (gh)a(h^{-1}g^{-1})$ $\implies$ $g(hah^{-1})g^{-1}$. Now, I am confused because $hah^{-1}$ $\in A$, but it is not necessarily equal to $a$ which means, if I say $hah^{-1} = b \in A$ then I get, $gbg^{-1} \in A$. What I am not sure is I started with the assumption that the element $a$ must be the same. What am I misunderstanding here?

Juan L.
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Four Seasons
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2 Answers2

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I can only presume the subset $G$ should have been written as $$G = \{m \in M: mam^{-1} \in A \quad \forall a \in A\}.$$ Then arguing as you do you run into no barriers, as you correctly identified that $hah^{-1} \in A$, so then so too $ghah^{-1}g^{-1} \in A$, and $G$ is closed.

I'll leave you to consider why $G$ is closed under taking inverses.

DanLewis3264
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  • Thank you for your reply. Could you see if my work for closure under inverses makes sense? So I took $m \in M$ such that $mam^{-1} \in A \forall a \in A$. Since $A$ $\leq$ $M$, $mam^{-1} \in A \forall a \in A$ $\implies$ $(mam^{-1})^{-1}$ $\in$ $A$. But that does not lead me anywhere, does it? How should I approach this? – Four Seasons Mar 14 '19 at 16:28
  • Does it not? $(mam^{-1})^{-1} = m^{-1}a^{-1}m$, so if this is in $A$ for all $a \in A$ (and necessarily all $a^{-1} \in A$, since $A \leq M$), then doesn't that say $m^{-1} \in M$, which is precisely what you want to show? – DanLewis3264 Mar 14 '19 at 16:42
  • But isn’t $(mam^{-1})^{-1} = ma^{-1}m^{-1}$ using the inverse rule for products of elements in a group? – Four Seasons Mar 14 '19 at 16:47
  • Sorry, yes, you're quite right, let me rethink. – DanLewis3264 Mar 14 '19 at 16:58
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    If $m \in G$ then $mam^{-1} \in A \quad \forall a \in A$. Now $m^{-1}(mam^{-1})m \in A$ for all $a \in A$ (you will need to convince yourself why all $mam^{-1} \in A$ give all $a \in A$), so $m^{-1} \in G$ as required. – DanLewis3264 Mar 14 '19 at 17:03
  • Just a quick question. So I see why $mam^{-1} \in A$ gives all $a \in A$ cause $A$ is a subgroup of $M$ and since $M$ is a group and is closed, $mam^{-1} \in M$. And since all elements of M are obtained that way, all elements of A are also obtained. What I am struggling to see is: Is it possible for $mam^{-1}$ to produce elements other than that in $A$? – Four Seasons Mar 14 '19 at 17:22
  • Not if $m \in G$! If $m \in M \setminus G$ then yes – DanLewis3264 Mar 14 '19 at 17:23
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$G$ is called the normalizer of $A$. You are on the right track.

If $g, h \in G$, then $gAg^{-1}=A$ and $hAh^{-1}=A$. So, $$ghA(gh)^{-1}=ghAh^{-1}g^{-1}=gAg^{-1}=A.$$ Thus, $gh \in G$.

If $g \in G$, then $gAg^{-1}=A$ and thus $A=g^{-1}Ag$. So, $g^{-1} \in G$.

Juan L.
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