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Let $H$ be a subgroup of a group $G$. $H$ is normal iff $aH=Ha$ for every $a \in G$

I'm having trouble with this proof both ways actually.

Assume $H$ is a subgroup of a group $G$ and $H$ is normal. Since $H$ is normal then for $h \in H$ there exists $aha^{-1} \in H$ where $h=aha^{-1}$. Then:

$$h=aha^{-1}$$

Multiply the right by a, we have:

$$ha=ah$$ and since $h \in H$ then:

$$Ha=aH$$ i'm not sure if I can just jump from $ha=ah$ to $Ha=aH$. I feel like I'm missing a step.

Any ideas?

Also, I know I have to show if $aH=Ha$ then H is normal but I think I can do that on my own if I know how to do the first part of the proof.

Mark
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1 Answers1

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You have to be weary of the definition of $aHa^{-1}=H$ for $H$ normal. It does not mean $aha^{-1}=h$. It means there is $h'\in H$ such that $aha^{-1}=h'$. The first case means $h$ is central, or commutes with all $a\in G$. Now suppose $H$ is normal. Then $aHa^{-1}=H$, or $aha^{-1}=h'$. Thus $ah=h'a$, or $aH=Ha$. The other direction is similar.

Moya
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