This abstract algebra question explores the connection between the commutativity of a group and a special subgroup called the commutator subgroup.
Step 1: Understanding the Question:
We need to find the specific property of the commutator subgroup that is equivalent to the group being Abelian (commutative).
Step 2: Key Formula or Approach:
We must use the definitions of an Abelian group, a commutator, and the commutator subgroup to establish the connection.
Step 3: Detailed Explanation:
Definition of an Abelian Group: A group \(G\) is Abelian if and only if \(ab = ba\) for all elements \(a, b \in G\).
Definition of a Commutator: The commutator of two elements \(a\) and \(b\) is defined as \([a,b] = aba^{-1}b^{-1}\). The commutator measures how much \(a\) and \(b\) fail to commute. If \(ab=ba\), then we can rewrite this as \(aba^{-1} = b\), and further as \(aba^{-1}b^{-1} = e\), where \(e\) is the identity element. So, \(a\) and \(b\) commute if and only if their commutator \([a,b]\) is the identity.
Definition of the Commutator Subgroup: The commutator subgroup, denoted \(G'\) or \([G,G]\), is the subgroup generated by all the commutators in \(G\).
Connecting the Concepts: If the group \(G\) is Abelian, then by definition, \(ab=ba\) for all \(a,b \in G\). This implies that every commutator \([a,b]\) is equal to the identity element \(e\). The subgroup generated by a set containing only the identity element is the smallest possible subgroup, \( \{e\} \), which is called the trivial subgroup.
Conversely, if the commutator subgroup is trivial (\(G' = \{e\}\)), it means all commutators are equal to \(e\), which implies \(ab=ba\) for all elements, so the group is Abelian.
Step 4: Final Answer:
A group \(G\) is Abelian if and only if its commutator subgroup is trivial.