Question:Normal subgroup definition inequality confusion

See also Question:Normal subgroup multiple definitions.

Q: '''One definition I saw said that $$N$$ is normal in $$G$$ if $$gNg^{-1} \le N$$ for all $$g \in G$$. Another definition used $$gNg^{-1} = N$$. Why are these equivalent?'''

A: This is a very good and somewhat tricky question. The $$=$$ formulation seems stronger, but it turns out to be equivalent. Interestingly, it is not true that if, for a particular element $$g$$ and a subgroup $$N$$, $$gNg^{-1} \le N$$, then $$gNg^{-1} = N$$. Rather, it is the fact that this is true for every $$g$$ that matters.

This is because restriction of automorphism to subgroup invariant under it and its inverse is automorphism. Now, if $$N$$ is normal in $$G$$ by the $$\le$$ definition, we not only have $$gNg^{-1} \le N$$, we also have that $$g^{-1}Ng \le N$$. Thus, $$N$$ is invariant both under conjugation by $$g$$ and under conjugation by $$g^{-1}$$, which are inverse operations of each other. This forces $$gNg^{-1} = N$$. (To see this without using the fact quoted above, note that $$g^{-1}Ng \le N$$, implies, via left multiplication by $$g$$ and right multiplication by $$g^{-1}$$, that $$N \le gNg^{-1}$$. Combined with $$gNg^{-1} \le N$$, we obtain $$N = gNg^{-1}$$).

Note also that the use of the $$\le$$ symbol, as opposed to the $$\subseteq$$ symbol, already encodes the information that conjugation by $$g$$ is an automorphism, hence the image of $$N$$ is a subgroup. If you read this definition before these ideas were introduced, you may have seen the notation $$gNg^{-1} \subseteq N$$. The same ideas apply with the set notation.