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Let $G$ be a group, let $A, B$ be subgroups of $G$, and assume $A \le N_G(B)$.

My question comes from reading Dummit and Foote, $\S 3.3$: The Isomorphism Theorems. We are proving the Second/Diamond Isomorphism Theorem, and they state in the proof:

Since $A \le N_G(B)$ by assumption and $B \le N_G(B)$ trivially, it follows that $AB \le N_G(B)$ i.e. $B$ is a normal subgroup of the subgroup $AB$

The italicized implication has me confused. I believe that I have sorted it out with the following conclusion, but I would like to check:

If a group $C \le N_G(B)$, then $B \trianglelefteq C \iff B \subseteq C$

Thus I cannot make a similar claim about $A$, referring back the the title, because I do not know that $B$ is a subgroup of $A$, but I do know that $B$ is a subgroup of $AB$?

Thanks.

1 Answers 1

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$$x\in N_G(H)\Longleftrightarrow H^x= H$$

and from this it follows that

$$B\leq A\leq N_G(H)\Longrightarrow \,\forall\,a\in A\,\,,\,H^a\subset H\Longrightarrow B\triangleleft A$$

Since $\,B\leq AB\leq N_G(H)\Longrightarrow B\triangleleft AB\,$

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    I think you should change the first line since $H^x \subseteq H$ does not imply $x \in N_G(H)$. The definition of normalizer is those elements of $x$ for which $H^x = H$. The set of those elements for which $H^x \subseteq H$ is closed under the group operation but is not necessarily closed under inverses (unless, for example, we assume $H$ is finite).2012-11-03
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    Interesting comment @m.k. Could you provide an example as to where this might go wrong when consider an infinite group? I'm trying to see if this might go wrong for $D_\infty$, the infinite dihedral group, as this seems to be where many of our counterexamples have come from2012-11-03
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    @jmi4: There are two examples given in [this question](http://math.stackexchange.com/questions/107862/conjugate-subgroup-strictly-contained-in-the-initial-subgroup/107874#107874). Finding an example of this is the same as finding $x$ and $H$ such that $xHx^{-1} \subseteq H$ but $xHx^{-1} \neq H$.2012-11-03