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I am given the left coset multiplication rule which is :

$$aHbH=abH$$

Taking representatives $x \in aH $ and $a^{-1} \in a^{-1}H$ $$xHa^{-1}H=xa^{-1}H$$

Take representatives $a\in aH$ and $a^{-1}\in a^{-1}H$ Then

$$aHa^{-1}H=H$$

If, multiplication is well defined then, $xa^{-1}H=H \implies xa^{-1}=h \ \ for \ some \ h\in H \implies x=ha \implies aH \subseteq Ha$

Now in order to prove the containment in other direction what should I do? I can easily do the exact same thing, but I'm not given a definition for right coset multiplication - I don't know how to multiply right cosets? Should I act as if I was given the such property? If I should, why? If shouldn't how ?

augustoperez
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    You've shown $aHa^{-1}H=H$ and that immediately gives $aHa^{-1}=H$ which is a definition of a subgroup being normal. Or if you prefer $aH=Ha$, you simply postmultiply $a$ to get it. – daruma Apr 22 '17 at 09:34

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I prefer the definition of a normal subgroup as the kernel of a homomorphism. Then you're told that $\phi: a \mapsto aH$ is a homomorphism (because coset multiplication is well-defined); its kernel is just $H$, so you're done.

Patrick Stevens
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