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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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    $\begingroup$ 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 $G$. $\endgroup$ – Rankeya Nov 7 '11 at 3:17
  • $\begingroup$ i don't think a normal subgroup has to have order 2 though $\endgroup$ – Scharfschütze Nov 7 '11 at 3:20
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    $\begingroup$ I don't mean in general. Just in cases when it applies. $\endgroup$ – Rankeya Nov 7 '11 at 3:21
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    $\begingroup$ "In particular, if p is the smallest prime dividing the order of G, then every subgroup of index p is normal." $\endgroup$ – The Chaz 2.0 Nov 7 '11 at 3:41
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    $\begingroup$ 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. $\endgroup$ – Ragib Zaman Nov 7 '11 at 3:46
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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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    $\begingroup$ "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! :) $\endgroup$ – Anna B Nov 7 '11 at 3:36
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    $\begingroup$ @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. $\endgroup$ – Arturo Magidin Nov 7 '11 at 3:38
  • $\begingroup$ @ArturoMagidin Can I know why 3. will tell us that N is normal? $\endgroup$ – user10024395 Nov 17 '14 at 13:51
  • $\begingroup$ Suppose 3. holds. Then consider a simple case $(y_1 y_2) N (y_1 y_2)^{-1} = (y_1 y_2) N (y_2^{-1} y_1^{-1}) = y_1 N y_1^{-1} = N$. So if $y_i$ generate $G$, then any $g N g^{-1}$ can be written as $(y_1 y_2 \dots y_k)N(y_1 y_2 \dots y_k)^{-1}$ and we apply this argument. $\endgroup$ – eatfood Oct 12 '19 at 16:53
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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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  • $\begingroup$ (+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 $A_n.$ $\endgroup$ – Ragib Zaman Nov 7 '11 at 3:28
  • $\begingroup$ I've thought of isomorphism too, but i think it's often not easy to find such an isomorphism. $\endgroup$ – Scharfschütze Nov 7 '11 at 3:38
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    $\begingroup$ @Scharfschütze: You want a homomorphism, not an isomorphism... if it's an isomorphism, you'll only prove the trivial subgroup is normal. $\endgroup$ – Arturo Magidin Nov 7 '11 at 3:39
  • $\begingroup$ @arturo magidin: sorry I meant homomorphism. But isn't it true that G/N is isomorphic to the image of this homomorphism? $\endgroup$ – Scharfschütze Nov 7 '11 at 3:43
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    $\begingroup$ 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. $\endgroup$ – KCd Nov 7 '11 at 3:46
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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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  • $\begingroup$ Can you explain why? $\endgroup$ – The Coding Wombat Oct 31 '18 at 12:17

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