Left coset is generated by the element of G, but not H

Question

left coset of H in G is a subset of G of type aH for some element a of G. Why does nobody requires element a not to belong to H itself?

Answer 1

Why not? If $a\in H$, then $aH=H$, which is indeed a coset.

Answer 2

The (left) cosets of $H$ are the sets $aH = \{ ah : h \in H \}$, and two cosets are equal if they are equal as sets, so maybe $aH = bH$ even if $a \neq b$.

With this in mind you can see that $\forall g \in H$, $gH = \{ gh : h \in H \} = H$. So every element of $H$ makes the same left coset of $H$, which is just $H$ itself.

For different elements outside of $H$, we will get other cosets that are different from $H$ itself, although usually there will be multiple elements that will give you the same coset of $H$. In fact, the only time that every element of $G$ will give a different left coset of $H$ will be if $H$ is the trivial subgroup.

And the direct answer to your question is that, as mentioned, $H$ itself is a perfectly valid coset. In fact, when $H$ is normal, within the quotient group $G/H$, the coset $H$ is the identity element, and within the original group, the subgroup $H \trianglelefteq G$ is the kernel of the canonical homomorphism $\varphi: G \to G/H$