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I've been self studying abstract algebra from these notes. I've ran across the following Lemma: alt text

The confusing part is that $X$ is a "any subset" not any "subgroup", and that the term "normal set" appeared in place where I expected "normal subgroup". The definition of a "normal set" has not been provided and I couldn't find it with a quick Googling.

So the question is

What is a "normal set" and why is the proof of that Lemma obvious ?

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My guess is that it is a subset closed under conjugation (e.g. take the definition of a normal subgroup and remove the subgroup conditions) but I don't think this is standard. In any case I hope the proof is clear now. – Qiaochu Yuan Oct 8 '10 at 14:16
Yes, from that I could prove $M=\cup_{g\in G} gXg^{-1}$ is normal given that G is a group. But I have no clue how to prove it is the smallest normal set containing $X$... – M. Alaggan Oct 8 '10 at 14:35
If Y is a normal set containing X, then it must contain the conjugates of every element of X. – Qiaochu Yuan Oct 8 '10 at 14:38
And in this case it ONLY consists of the conjugates, so it is minimal. Thanks a lot :) – M. Alaggan Oct 8 '10 at 14:43
up vote 1 down vote accepted

A normal set is a set $X$ for which $g^{-1}xg\in X$ for every $x\in X,g\in G$. It's just like the normality condition for subgroups, except that $X$ doesn't have to be a subgroup.

We can see that $\bigcup_{g\in G}g^{-1}Xg$ must be the smallest normal set containing $X$ quite easily.

Proceed algorithmically. Begin with $N=X$. We want $g^{-1}xg\in X$ for every $x\in X,g\in G$, so for each $x\in X$, add $g^{-1}xg$ to $N$. Repeat this process until all such elements are included in $N$. Then $N$ is the smallest normal set of $G$ containing $X$.

Of course, $\langle g^{-1}Xg : g \in G \rangle$ would be the smallest normal subgroup of $G$ containing $X$.

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