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If $H <G$, then $H$ is called normal if $gHg^{-1} \subset H$ for all $g\in G$.

Why is the definition not

$gHg^{-1} = H$? We are not assuming commutivity for $G$.

Take $gH = Hg$, multiply both sides by $g^{-1}$ and we get $gHg^{-1} = Hgg^{-1}$. Why is $Hgg^{-1} \neq H$ always?

i.e, why do we use a $\subset$ instead of equality?

complexLost
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1 Answers1

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These two conditions are actually equivalent. Notice that the subset is a "subset or equal" and not a "strict subset". Clearly if $gHg^{-1}=H$ then $gHg^{-1} \subset H$.

Now suppose that $gHg^{-1} \subset H$ for all $g$.

Consider $h \in H$. Then, since $g^{-1} \in G$, it follows that $g^{-1}hg=h'$, where $h' \in H$. Thus, since $h=(gg^{-1})h(gg^{-1})=g(g^{-1}hg)g^{-1}=gh'g^{-1}$, it follows that $h \in gHg^{-1}$. Thus, $H \subset gHg^{-1}$, and so $H=gHg^{-1}$.

Some people prefer the subset version because it gets at the equivalent definition of "for all $g\in G$ and $h \in H$, $ghg^{-1} \in H$ ". That is, $H$ is closed under conjugation by every element in $G$.

ASKASK
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  • Why is $g^{-1}hg \in H$ ? Sorry if this is obvious. So $g^{-1} \in G$, then $g^{-1}H$ is a left coset which is always a subset of $G$. I guess it is true that if you take a left coset then do the operation on the righthand side, that is also in $G$? – complexLost Dec 04 '15 at 05:58
  • We are given that for all values of $g \in G$, it is true that $gHg^{-1} \subset H$. In particular, since $g^{-1} \in G$, it follows that $g^{-1}H(g^{-1})^{-1}=g^{-1}Hg \subset H$. Thus, for all values of $h$, $g^{-1}hg \in H$. – ASKASK Dec 04 '15 at 06:01
  • All we are doing is replacing $g$ with $g^{-1}$, which we can do because $g^{-1} \in G$ – ASKASK Dec 04 '15 at 06:01
  • Oh right we were assuming that was true. For some reason I thought you meant that was true in general. I understand now. Thank you. – complexLost Dec 04 '15 at 06:03
  • except you're using the strict subset sign =/ $A \subseteq B$ can have $A=B$ but $A \subset B$ cannot as that is a strict subset – Zelos Malum Dec 04 '15 at 10:47
  • @zelos many people use the sign to mean both strict and non strict. I simple kept consistent with whatever book he is learning from. Clearly in this case it was meant to be non strict – ASKASK Dec 04 '15 at 11:51
  • Which makes no sense as $\subseteq$ is the one used for it. – Zelos Malum Dec 04 '15 at 11:53
  • It is all just notation. MANY courses/professors/textbooks (clearly including the one OP uses) use the notation $\subset$ to mean "subset or equal" and use $\subsetneq$ to explicitly mean "strict subset". I think the fact that you are down voting my answer because it uses notation different from what you're used to is highly immature. – ASKASK Dec 04 '15 at 12:02
  • See http://math.stackexchange.com/a/428190/136368 for a discussion of this notation – ASKASK Dec 04 '15 at 12:05
  • I would like you to please undo your downvote since this answer is correct and answers the question. – ASKASK Dec 04 '15 at 14:46
  • @ASKASK: I upvoted but actually I might be in favor of the $\subseteq$ version. In fact, I use $\subset$ only when $\subsetneq$ is not needed, but will always occur (e.g. $K \subset \mathbb R^n$ compact) – Cloudscape Dec 04 '15 at 18:57