(Corrected in Wikipedia) Every subsemigroup is in its own normalizer

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Let $G$ be a group and $S \subseteq G$. We define the normalizer of $S$ as $N(S) := \{ n \in G : nS = Sn\}$

According to Wikipedia;

If $S$ is a subsemigroup of $G$, then $N(S)$ contains $S$.

But I can't prove it. Everything that $S$ being a subsemigroup tells me is that $\forall s \in S,\ sS \subseteq S$ and $Ss \subseteq S$

How would one derive that $\forall s \in S,\ sS = Ss$ ? If S were a subgroup or finite, it would be trivial.

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This is not true. Consider the free group generated by two symbols $a,b$. Let $A$ be its subsemigroup generated by $a,b$. Now $a\in A$, but $aA$ consists of all strings in $a,b$ starting with $a$, while $Aa$ consists of all strings in $a,b$ terminating with $a$. They are obviously different.