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I think in most situations(for example, in $S_n$ or $D_n$), proving by definition is too complicated because you have to calculate $gng^{-1}$ for every $n$ in $N$ and $g$ in $G$. To prove that all the left cosets are also right cosets is also too complicated because you have to find all those cosets. I wonder if there's a way to do this without having to calculate everything by hand.

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    This result has often helped me: If $N$ is a subgroup of a group $G$ such that $[G:N]=2$, then $N$ is normal in $G$.2011-11-07
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    i don't think a normal subgroup has to have order 2 though2011-11-07
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    I don't mean in general. Just in cases when it applies.2011-11-07
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    Also by $[G:N]$ I mean the index of $N$ in $G$. Not the order of $N$.2011-11-07
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    "In particular, if p is the smallest prime dividing the order of G, then every subgroup of index p is normal."2011-11-07
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    The Chaz gave a nice generalization. Here's three more that should incorporate into their answers: $H$ is normal if it is the union of conjugacy classes in $G.$ $H$ is normal if the commutator $[H,G] \subseteq H.$ If $G$ is a nilpotent group, every maximal subgroup is normal.2011-11-07
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    Oh wow. Here's a [nice](http://groupprops.subwiki.org/wiki/Proving_normality) website!2011-11-07

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There are a number of ways in which the work can be shortened.

  1. If you can come up with a homomorphism whose kernel is precisely $N$, then this guarantees that $N$ is normal. This is often the case.

  2. It suffices to check a generating set for $N$. That is, if $N=\langle X\rangle$, then $N$ is normal in $G$ if and only if $gxg^{-1}\in N$ for every $x\in X$. For instance, this makes proving that the subgroup generated by all $m$ powers is normal easy.

  3. It suffices to check a generating set for $G$ and its inverses. That is, if $G=\langle Y\rangle$, and $yNy^{-1}\subseteq N$ and $y^{-1}Ny\subseteq N$ for all $y\in Y$, then $N$ is normal.

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    "If you can come up with a homomorphism whose kernel is precisely N, then this guarantees that N is normal. This is often the case." - This is _always_ the case! :)2011-11-07
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    @AnnaB: It's always the case that a homomorphism exists. It is not always the case that you can *come up* with such a homomorphism *without* knowing that $N$ is normal, though.2011-11-07
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    @ArturoMagidin Can I know why 3. will tell us that N is normal?2014-11-17
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If your subgroup has index 2, then it is always normal (because whether you consider left or right cosets, there are only these 2: the subgroup itself, and the rest of the elements).

Another way (maybe the best way) is to show that the subgroup is the kernel of a homomorphism having the group as its domain.

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    (+1) It is interesting to note that all 3 collected ways (definition, index 2, kernel) give quick and easy proofs that $A_n$ is normal in $S_n.$ We should probably expect from any decent method for proving normality that it works nicely on $A_n.$2011-11-07
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    I've thought of isomorphism too, but i think it's often not easy to find such an isomorphism.2011-11-07
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    @Scharfschütze: You want a homomorphism, not an isomorphism... if it's an isomorphism, you'll only prove the trivial subgroup is normal.2011-11-07
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    @arturo magidin: sorry I meant homomorphism. But isn't it true that G/N is isomorphic to the image of this homomorphism?2011-11-07
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    @arturo magidin: anyway, it's not easy to find such a homomorphism!2011-11-07
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    @Scharfschütze: Yes, and yes. (That's why in my answer I say "this is *often* the case"; i.e., it is often the case that one can find such homomorphisms, but not indeed not always). However, with more experience, you will continue to grow your internal "album of groups" (groups that you are familiar with and can call upon at need), which will also help you come up with more possible homomorphisms when you encounter a new group.2011-11-07
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    One basic case where the kernel method does not work is the commutator subgroup. Of course *after* you have checked that subgroup is normal you can come up with a homomorphism with it as the kernel, but I've never seen a proof that the comm. subgroup is normal using group homomorphisms.2011-11-07
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    @arturo magidin: yeah i know... but i'm still very unexperienced ==2011-11-07
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That would depend on the problem. I believe the following properties are most useful.

A subgroup $N$ of $G$ is normal iff one of the following is true:

  1. For every $g\in G$ and $n\in N$, $gng^{-1}\in N$.
  2. For every $g\in G$, $gNg^{-1}\subseteq N$.
  3. For every $g\in G$, $gNg^{-1}=N$.
  4. Every left coset of $N$ is a right coset of $N$.
  5. The product of two right cosets of $N$ is again a right coset of $N$.
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If you know a subgroup of a particular order is the ONLY subgroup of that order, then you know it's normal. I know that's a unique case, but just another tool to keep in mind.

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    Can you explain why?2018-10-31