Since aman=anama^ma^n=a^na^maman=anam for every am∈⟨a⟩a^m\in\langle a\rangleam∈⟨a⟩ and an∈⟨a⟩a^n\in\langle a\ranglean∈⟨a⟩, we obtain that ⟨a⟩\langle a\rangle⟨a⟩ is commutative.
Answer: commutative
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