A commutative group G is simple if and only if-
A group G is called a commutative group (or abelian group) if its group operation is commutative, meaning for any elements \(a, b \in G\), we have \(ab = ba\).
A group G is called a simple group if its only normal subgroups are the trivial subgroup \(\{e\}\) (containing only the identity element) and the group G itself.
A key property of commutative groups is that every subgroup is a normal subgroup. This is because for any subgroup H and any element \(g \in G\), the conjugate subgroup \(gHg^{-1}\) is equal to \(gH g^{-1} = (g g^{-1}) H = eH = H\) since multiplication is commutative. Therefore, \(gHg^{-1} = H\), which is the definition of a normal subgroup.
Given that every subgroup of a commutative group G is normal, for G to be a simple commutative group, it must have no non-trivial proper normal subgroups. Since all subgroups are normal, this means G must have no non-trivial proper subgroups at all.
So, a simple commutative group G can only have \(\{e\}\) and G as its subgroups.
Let the order of the group be \(O(G) = n\). We need to find the condition on \(n\) such that G has no subgroups other than \(\{e\}\) and G.
Consider the possible orders of subgroups by Lagrange's Theorem. Lagrange's Theorem states that the order of any subgroup of a finite group divides the order of the group.
Combining these points, a commutative group G is simple if and only if its order \(O(G)\) is a prime integer.
For a commutative group G to be simple, its order must not be a composite number or 1 (a group of order 1 is trivially simple, but often prime order is the defining case as it's the smallest non-trivial simple group). The order must be a prime integer.
The condition is \(O(G) = n\), where \(n\) is a prime integer.
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