Let be $H$ a subgroup. Show $Ha=H$ if and only if $a$ belongs to $H$.
Here what I have understand of the proof: $a$ must belong to $Ha$ because $ea=a$ where $e$ is the identity in $H$. So the only way $H=Ha$ is that a belongs to $H$.
Is that correct? Now suppose that a belongs to H. for every element h in H, $h=(ha^{-1})a$. This implies that h belongs to Ha so also a belongs to Ha. This implies H=Ha.
$\endgroup$52 Answers
$\begingroup$If $a\in H$, then every element of $H$ can be written in the form $(ha^{-1})a\in Ha$ and due to closure, $Ha=H$. Contrarily, if $Ha=H$ then $a=ea\in Ha=H$.
$\endgroup$3$\begingroup$Let $a\in G$. Then
\begin{align} Ha=H &\iff Ha=He \\ &\iff ae^{-1}\in H \\ &\iff a\in H. \end{align}
This is due to the more general theorem that $Hx=Hy$ if and only if $xy^{-1}\in H$.
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