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(1.) Not querying proofs or formality. I do this in my other question.

Normal Subgroup Test says H is normal in G $\iff gH{g}^{-1}\subseteq H$ for all $g \in G$.
What's the intuition of this subset 'equation'?

(2.) Intuitively, how's a subset equation equivalent to 1,2,3 overhead?

p. 6 http://www.millersville.edu/~bikenaga/abstract-algebra-1/normal/normal.html:
If you know a subgroup is normal, $gHg^{-1} = H$ is the "stronger" form. On the other hand, if you're trying to prove that a subgroup H is normal in a group G, all you need to show is that $gHg^{-1} \subseteq H$ for all $g \in G$ (as opposed to equality).

(3.) How is equality a "stronger" form than the subset form, when these two are proven as equivalent? Stronger $\neq$ equivalent?

  • 1
    $\begingroup$ Instead of "querying", try "asking"; replace "overhead" with "above". $\endgroup$ – Namaste Apr 28 '14 at 14:14

I hope that this answers (partially) your question:

For a specific $g\in G$ the statement $gHg^{-1}=H$ is a stronger statement than $gHg^{-1}\subset H$. The first implies the second but the opposite is not necessarily true.

However the statements $\forall g\in G\; gHg^{-1}\subset H$ and $\forall g\in G\; gHg^{-1}=H$ are equivalent.

This because from the first we can deduce that also $H=g\left(g^{-1}Hg\right)g^{-1}\subset gHg^{-1}$ for each $g\in G$.


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