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Let $X$ be a set in $G$ and $G$ be a group. A normal set is a set $X$ for which $gxg⁻¹∈X$ for every $x∈X,g∈G$. It's just like the normality condition for subgroups, except that $X$ doesn't have to be a subgroup.

I hvae two questions:

(1) Can we define the normal set without $G$ being a group?

(2) The normal vector to a surface is a vector perpendicular to it. Does there is a relation between this notion and the notion of normal set?

drhab
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DER
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    $X$ is not just any set, it is a subset of $G$. Otherwise, $gxg^{-1}$ doesn't make sense. – fkraiem May 22 '14 at 08:50
  • @fkraiem: Yes, corrected. – DER May 22 '14 at 08:52
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    Firstly, your definition of normality is off slightly - you also need the map $x\mapsto gxg^{-1}$ to be a bijection. Secondly, how are you hoping to define $g^{-1}$? (Incidentally, if you take $G$ to be a group and don't want $X$ to be a subgroup then this is a decent question with a concrete "yes" answer.) – user1729 May 22 '14 at 08:52
  • @user1729: It is the invrse element of $g$. – DER May 22 '14 at 08:54
  • Yes, but if $G$ is not a group then you cannot guarantee inverses. – user1729 May 22 '14 at 08:55
  • @user1729: So the answer is NO. What about the second question. – DER May 22 '14 at 08:56
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    @DER Why do you think that they should be connected? – user1729 May 22 '14 at 08:57
  • @user1729: They have the same name and this is just an idea. – DER May 22 '14 at 08:59
  • @DER Sorry, of course. I'll think about it. – user1729 May 22 '14 at 08:59
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    @DER I have thought about it. I think you want to look up something called a "Lie algebra". I said earlier that the issue with the idea of a normal set is that you need inverses. So suppose you have inverses, so you also have an identity and a binary operation, but then you also want $G$ to not be a group. Therefore, you need the binary operation to not be associative. Lie algebras are vector spaces with an operation called a Lie bracket, and this bracket is not associative. Moreover, 3D vectors with the operation "cross product" form a Lie algebra. Score! This may be a red herring though... – user1729 May 22 '14 at 09:05
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    (Also, I should say that in semigroup theory there are certain notions which "mimic" normality, so you get things like $ag\overline{g}b=ab$ for all $a, b\in S$. So inverses are not strictly necessary for what you are after. I just think that you perhaps need to think about your question slightly more.) – user1729 May 22 '14 at 09:09
  • @user1729: Ok I will look for Lie Algebra. – DER May 22 '14 at 09:09
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    @user1729 If $gXg^{-1}\subset X$ is true for every $g\in G$ then the map $x\mapsto gxg^{-1}$ automatically is a bijection. Its inverse is the map $x\mapsto g^{-1}xg$. – drhab May 22 '14 at 09:21
  • @drhab: Yes, this property is satisfied. – DER May 22 '14 at 09:24
  • @drhab Touché . – user1729 May 22 '14 at 09:28
  • @drhab That's not correct: a conjugation map can restrict to a subset (a subgroup even) and then fail to be onto. See here. – anon May 22 '14 at 09:45
  • OP: (1) I imagine $\forall a\in G,aX=Xa$ would work in a set $G$ equipped with a binary operation. This is equivalent to normality if $G$ is a group, but still makes sense if it isn't. (2) No, orthogonality in geometry and conjugation invariance in a symmetry group aren't related any more than any other randomly picked pair of mathematical concepts. – anon May 22 '14 at 09:50
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    @seaturtles No, drhab is correct. The key point is that it holds for all $g\in G$. The example you give - and indeed any example - does not hold for every $g\in G$. Indeed, $gHg^{-1}<H\Rightarrow H<g^{-1}Hg$. – user1729 May 22 '14 at 10:19
  • @user1729 Ah, thanks. I wasn't incorporating the for all. – anon May 22 '14 at 15:29
  • @seaturtles Yes, I made the same mistake in my initial comment! – user1729 May 23 '14 at 12:09

2 Answers2

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Your definition of normal subsets $X$ of group $G$ is absolutely valid. One can easily see that a subset is normal iff it is the union of conjugacy classes (a conjugacy class of an element $x \in G$ is the set $\{g^{-1}xg: g \in G\}$, which is the equivalence class of $x$ under the equivalence relation conjugation). Conjugacy classes are important in groups for several reasons beyond the scope of this answer. And with them, one can actually "construct" normal sets. Observe that complements $G-N$ of normal subgroups $N$ of $G$ are normal subsets.

In a similar vein one can define the centralizer $C_G(X)=\{g \in G:g^{-1}xg=x$ for all $x \in X\} $ and normalizer of a set $X$, $N_G(X)=\{g \in G:g^{-1}Xg=X\}$.

Nicky Hekster
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A congruence in a monoid $M$ is an equivalence relation $\equiv$ in $M$ that is compatible with the operation of $M$: $$ a \equiv b, a' \equiv b' \implies aa '\equiv bb' $$ The quotient $\overline M = M\,/\equiv$ is then a monoid.

It is easy to prove that, when $G$ is a group and $\equiv$ is a congruence in $G$, the equivalence class of $1$ is a normal subgroup $N$ of $G$ and the equivalence classes are the cosets of $N$. Conversely, if $N$ is a normal subgroup of $G$, then the relation defined by $a \equiv b$ if $a^{-1}b \in N$ is a congruence relation in $G$ whose equivalence classes are the cosets of $N$.

I don't think there is much to say about congruences in a monoid in general because given a submonoid $N$ of a monoid $M$, the cosets of $N$ in $M$ are not necessarily disjoint and so do not define equivalence classes.

lhf
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