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Let $H$ be a subgroup of $G$, and let $K = \{x \in G: xax^{-1} \in H \iff a \in H\}$.

Does this mean that $K$ is all of $G$? The statement seems to say nothing about $x$ itself. If there is no $a \in H$ that $x$ conjugates, then the antecedent is false so that statement is trivially true.

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I think this is a 'standard' interpretation of logical symbols: the test "$xax^{-1}\in H\Leftrightarrow a\in H$" has an implicit universal quantifier tying the free variable $a$ to its largest natural scope (here the bigger group $G$). So $x$ is in the set $K$, if and only if for all elements $a\in G$, $a$ is in $H$ exactly when $xax^{-1}$ is. IOW $$\forall\,a\in G: xax^{-1}\in H\leftrightarrow a\in H.$$ After all, the membership test must be a predicate that has a defined value for a given 'candidate element' $x$. For that reason it is necessary to use the universal quantifier to close it. – Jyrki Lahtonen Jan 16 '12 at 9:08
up vote 3 down vote accepted

The statement $xax^{-1} \in H \Longleftrightarrow a \in H$ for all $a \in H$ is logically equivalent to the equality of sets $xHx^{-1} = H$. That might help you understand $K$ better.

For an example where $K \neq G$, let $G = S_3$ (the group of permutations of a set with $3$ elements) and $H = \{e, (1,2)\}$ (the subgroup consisting of the identity permutation and the permutation interchanging $1$ and $2$). To see this, consider the permutation $x = (1,3)$ (interchanging $2$ and $3$). Then $xHx^{-1} = \{e, (2,3)\} \neq H$, so $x \notin K$. As an exercise, you should compute $K$ for this example.

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Thanks, I think this will help. I need to think about this. – Matt Gregory Jan 16 '12 at 2:51
Okay, I just realized that $H$ is already determined! I was thinking that $H$ was being defined at the same time as $K$, and that was really confusing. – Matt Gregory Jan 16 '12 at 3:05
Glad it is making more sense to you! The subgroup $K$ is called the normalizer of $H$ in $G$. It is the largest subgroup of $G$ in which $H$ is normal. That means: if $K' \subseteq G$ is any subgroup of $G$ which has the property that it contains $H$ as a normal subgroup, then $K'$ is a subgroup of $K$. You'll often see the notation $N_G(H)$ or just $N(H)$ for the normalizer of $H$ in $G$, instead of $K$. – Michael Joyce Jan 16 '12 at 3:27

"If there is no $a \in H$ that $x$ conjugates" makes no sense. You can use $x$ to conjugate any element $a$; the question is what the result will be, and the statement says that $K$ contains the elements $x$ of $G$ such that conjugating $a$ with $x$ yields an element of $H$ iff $a$ is in $H$.

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$$xax^{-1} \in H \iff a \in H$$

means that if $a$ is in $H$, then $xax^{-1}$ is in $H$, and if $a$ is an element of $G$ such that $xax^{-1}$ is in $H$, then $a$ is also in $H$. So $x$ is in $K$ if the above happens. To contrast, $x$ is not in $K$ if there exists $a\in H$ such that $xax^{-1}$ is not in $H$, or if there exists $a\in G\setminus H$ such that $xax^{-1}$ is in $H$.

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But if $xax^{-1}$ is not in $H$, doesn't that mean that $a$ would not be in $H$? – Matt Gregory Jan 16 '12 at 2:45
@Matt: If $x$ is in $K$, then yes that would be implied by definition of $K$. But not in general; that is why the definition of $K$ gives you a condition that $x$ must satisfy to be in $K$. If you are familiar with normal subgroups, consider the following. If $H$ is not a normal subgroup, then there exists $a\in H$ and $x\in G$ such that $xax^{-1}$ is not in $H$. In that case, $x$ is not an element of $K$. On the other hand, if $H$ is a normal subgroup, then $K=G$. – Jonas Meyer Jan 16 '12 at 2:52

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