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Let $G$ be a group and $N$ be a subgroup of $G$. Then the following statements are equivalent:

The subgroup $N$ is normal in $G$.
For all $g\in G$, $gNg^{-1}\subseteq N$.
For all $g\in G$, $gNg^{-1}=N$
I don't understand why we need the second condition. We could've just put the 3rd one and the first one. What was the need for the second one? it just seems that it's redundant.

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    $\begingroup$ When you want to prove something, It is easy to prove the weaker condition. When you want to use something, It is better to use the stronger condition. So when you want to prove some group is normal, you may prove the second condition. When you know some group is normal and want to use that property, you may want the third one $\endgroup$
    – Koobe
    Apr 15 at 23:42
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    $\begingroup$ It is always useful to have multiple ways of characterizing something. Here we define normal subgroup one way, but we establish that there are multiple conditions that are also equivalent; in some situations, one of the conditions might be easier to use/verify, in others a different one might be. In general, it is easier to verify weaker conditions, and easier to use stronger ones, as noted by Koobe. Note that we are saying "These statements are equivalent"; this is not a definition, this is a theorem that tells you each of the three statements implies the other two. $\endgroup$
    – Arturo Magidin
    Apr 15 at 23:58
  • $\begingroup$ Think of it as an important lemma free of charge included in the definition. Definition: $N$ is normal iff $gNg^{-1}=N$ for all $g$. Lemma: If $gNg^{-1}\subseteq N$ for all $g$, then $N$ is normal. $\endgroup$
    – Vercassivelaunos
    Apr 16 at 0:05
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    $\begingroup$ Just to be clear: it is possible to have a subgroup which properly contains a conjugate of itself. See this question for a simple example. Thus, conditions two and three are not obviously redundant (if that's what you were thinking...it's not clear from your post why you thought these definitions were redundant). $\endgroup$
    – lulu
    Apr 16 at 0:48
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    $\begingroup$ In a specific case, where you know $G,N,$ it will be easier to prove the second one than it is the third one. $\endgroup$
    – Thomas Andrews
    Apr 16 at 0:51

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So that you don't need to prove the reverse inclusion each time you want to test whether a group is normal.

That is, if $N$ is a subgroup of $G$, and I show that for all $n\in N$, $g\in G$, we have that $gng^{-1}\in N$, then $N$ is normal to $G$ automatically.

If we only relied on the third condition, we would have to further prove that for all $n\in N$, $g\in G$, $n\in gNg^{-1}$. Due to the statement given, we see that we automatically get this from the second condition.

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