Normalizer is subgroup of G

Asked May 07 '16 at 14:42

Active May 07 '16 at 15:15

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In my textbook, the definition of normalizer is

$N_G(A)=\{g\in G \mid gAg^{-1} \subseteq A\}$

Many textbooks use this definition.

$N_G(A)=\{g\in G \mid gAg^{-1} = A\}$

I know that normalizer is subgroup of $G$.

There is no problem when $A$ is finite. Two things are the same.

But if $A$ is infinite, I know that there exists example such that $ gAg^{-1}$ is strictly in $A$.

In that case, it can be shown that for $g \in N_G(A)$, $g^{-1}\in N_G(A)$?

edited May 07 '16 at 15:15

Andreas Caranti

asked May 07 '16 at 14:42

Pearl

1

do you mean that there is an example of $gAg^{-1}\subset A$ but $gAg^{-1}\neq A$? which is that? – janmarqz May 07 '16 at 14:51
2

The definition in the book wrong. For an example, see http://math.stackexchange.com/questions/107862 – Derek Holt May 07 '16 at 15:03
It is not clear exactly what the question is! – Derek Holt May 07 '16 at 15:48
@DerekHolt The question is if normalizer is defined as my textbook, it can be shown that it is subgroup of G. It is hard to show that it contains inverse element. – Pearl May 07 '16 at 15:55
It cannot be shown that it is a subgroup of $G$ because it is not always true. – Derek Holt May 07 '16 at 17:08

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