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Question

A commutative group G is simple if and only if-

The correct answer is O(G) = n, n is a prime integer

Understanding Commutative Simple Groups

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.

Commutative Group Properties

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.

Simple Commutative Group Condition

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.

  • A proper subgroup of G is a subgroup H such that \(H \neq G\).
  • A non-trivial subgroup is a subgroup H such that \(H \neq \{e\}\).

So, a simple commutative group G can only have \(\{e\}\) and G as its subgroups.

Order of a Simple Commutative Group

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.

  • If \(n\) is a prime number, say \(p\), then the only positive divisors of \(p\) are 1 and \(p\). By Lagrange's Theorem, the only possible orders of subgroups are 1 and \(p\). A subgroup of order 1 is the trivial subgroup \(\{e\}\), and a subgroup of order \(p\) must be the group G itself (since it has the same order as G). Thus, if \(O(G)\) is prime, G has no non-trivial proper subgroups. Since G is commutative, all subgroups are normal, so G is simple.
  • If \(n\) is a composite number, say \(n = ab\) where \(1 < a < n\). A commutative group of order \(n\) always contains an element of order \(d\) for every divisor \(d\) of \(n\) (this follows from the structure theorem for finite abelian groups, or more simply, by considering the existence of cyclic subgroups). If there is an element \(x\) of order \(a\), then the cyclic subgroup generated by \(x\), \(\langle x \rangle\), is a subgroup of order \(a\). Since \(1 < a < n\), \(\langle x \rangle\) is a non-trivial proper subgroup of G. As G is commutative, \(\langle x \rangle\) is a normal subgroup. Therefore, G is not simple if its order is composite.

Combining these points, a commutative group G is simple if and only if its order \(O(G)\) is a prime integer.

Conclusion on Commutative Simple Groups

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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Important Questions from Group & Subgroups

  1. Name the smallest non cyclic group.

  2. The generator of the group G = {a, a2, a3, a4, a5, a6 = e} is

  3. Let G = {1, -1, i, -i} be the multiplication group, and H = {1. -1} is a subgroup of G, then

  4. Let H be a subgroup of a group G and K be a normal subgroup of a group G, then

  5. If G is a group such that a2 = e for all a ∈ G, then G is:

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