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This question is a follow-up to this question. I am still not understanding how I can show more rigorously that $m^{-1}$ exists in G, in which case I have to prove that $m^{-1}am \in G \forall a \in A$. Does anyone have a more rigorous way of showing it that what the comments in the answer have? In particular, the answerer comments "If $m \in G$ then $mam^{−1} \in A \forall a\in A$. 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." How do I show that all elements in $A$ can be expressed as the product $mam^{-1}$?

Edit 1: Few things I have been asking myself recently is: Does the above proof mean exactly the same as proving $mAm^{-1} = A$ where $m \in M$? In that case, I have to show i) $mAm^{-1} \subseteq A$ and ii) $A \subseteq mAm^{-1}$. Number i) should be easy cause if $m \in M$, then $mam^{-1} \in A \forall a \in A$, which means $mAm^{-1} \subseteq A$. I do not know how to prove number ii) though.

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Four Seasons
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1 Answers1

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The definition of $G$ is $$ G=\{m\in M\mid mam^{-1}\in A\text{, for all }a\in A\} $$ The fact that $1\in G$ is obvious. Proving closure under products is easy: suppose $m,n\in G$ and $a\in A$; then, for $a\in A$, $$ (mn)a(mn)^{-1}=m(nan^{-1})m^{-1} $$ By definition, $b=nan^{-1}\in A$, so also $mbm^{-1}\in A$.

Now your question. If $m\in G$ and $a\in A$, then $mam^{-1}\in A$; therefore $mAm^{-1}\subseteq A$. You can't generally prove that $mAm^{-1}=A$, see Conjugate subgroup strictly contained in the initial subgroup?

The statement you have to prove requires more hypotheses, for instance that $A$ is finite, because in this case $mAm^{-1}$ has the same cardinality as $A$ and, being a subset thereof, they're equal.

If $a\in A\setminus mAm^{-1}$, then $m^{-1}am\in A$ would imply $a\in mAm^{-1}$: contradiction.

If we change the definition of $G$ to be $$ G=\{m\in M\mid mAm^{-1}=A\} $$ then $G$ is a subgroup of $M$.

egreg
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  • For me, this has raised more questions. How is the new definition of G that you have mentioned at the end of your answer going to help in the verification of the original question? If I did not get it wrong, are you suggesting that the fact that A is not mentioned in the question to be finite, there is no way to prove $mAm^{-1} = A$ and hence no way to prove that $m^{-1}$ exists in G? – Four Seasons Mar 19 '19 at 00:11
  • @Ufomammut The original question states a false theorem. – egreg Mar 19 '19 at 00:21
  • so would it be true if $A$ were mentioned to be finite? – Four Seasons Mar 19 '19 at 00:23
  • @Ufomammut Yes, that would be sufficient. – egreg Mar 19 '19 at 09:46