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Let $A$ and $B$ be two subgroups of the same group $G$. What does it mean for the subgroup $A$ to be normalized by the subgroup $B$?

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It means that $A \lhd \langle A,B \rangle$ is one way to say it. – Geoff Robinson Jul 29 '11 at 19:05

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It means that for every $b\in B$, $A^b = \{b^{-1}ab\mid a\in A\} = A$. That is, that $B$ is a subgroup of the normalizer of $A$ in $G$.

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could you please explain this statement: "If we want $AB=\{ab|a\in A, b\in B\}$ to be a subgroup of $G$, then we must have $A$ is normalized by $B$". – palio Jul 29 '11 at 19:27
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@palio: The statement is false; what is true is that if $A$ and $B$ are subgroups, then $AB$ is a subgroup if and only if $AB=\{ab\mid a\in A,b\in B\}$ is equal, as a set, to $BA=\{ba\mid a\in A,b\in B\}$. If $B$ normalizes $A$, then this is true: given $a\in A$, $b\in B$, then $ab = b(b^{-1}ab) = ba'\in BA$, so $AB\subseteq BA$, and $ba = bab^{-1}b = a''b\in AB $, so $BA\subseteq AB$. However, it is possible to have $AB=BA$ without having $B$ normalize $A$, so it is false that "we must have $A$ normalized by $B$". E.g., $G=S_3$, $A=\langle(1,2)\rangle, $B=\langle(1,2,3)\rangle$. – Arturo Magidin Jul 29 '11 at 19:39
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@palio: In my previous example, $A$ normalizes $B$, of course; but it is also possible to have $AB=BA$ and yet have neither $A$ normalize $B$, nor $B$ normalize $A$. The statement is simply false. – Arturo Magidin Jul 29 '11 at 19:39

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