Prove that every subgroup of an abelian group is a normal subgroup.
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The definition of a normal group is:
A group $H\leq G$ is a normal subgroup if for any $g\in G$, the set $gH$ equals the set $Hg$. Equivalently, you can also demand $H=gHg^{-1}$.
Now, take a subgroup $H$ of an abelian group $G$. Take any element $x\in gHg^{-1}$. By definition, this element must equal $ghg^{-1}$ for some $h\in H$.
Now you have $x=ghg^{-1}$, where $g\in G$ and $h\in H$. Can you prove $x\in H$?
5xum
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2Is it because since $G$ is abelian and $g,h \in G$ we have $gh=hg$? Because then I get $x=ghg^{-1}=hgg^{-1}=he=h \in H$, and so $gHg^{-1} \subseteq H$, which is one of the definitions of $H$ being a normal subgroup of $G$? – Cookie Mar 11 '19 at 02:27
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@Cookie Precisely. If $x=ghg^{-1}$, then $x=h$, and therefore, $x\in H$. – 5xum Mar 11 '19 at 07:32
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By definition, a normal subgroup $N\triangleleft G$ is one that $Ng=gN\ \forall g\in G$; that is, $\forall g\in G,n\in N\ \exists\,n'\in N: ng=gn'$ and $\forall g\in G,n\in N\ \exists\,n'\in N: n'g=gn$. For each of those, choose $g,n$ and let $n'=n$; because $G$ is abelian, you have the equation desired.
msh210
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Let $G$ be an abelian group and $H$ be subgroup of $G$. $H$ is also abelian group. We have that $ha=ah$ implies $aha^{-1}$ belong $H$, and hence that $H$ is a normal subgroup of $G$.
Mankind
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1You may want to check here to make your future (an present) posts easier to read. – Apr 21 '15 at 13:14