Proof: The Converse of “Every Cyclic Group is Abelian” is Not True
Problem
Show that the converse of the statement “Every cyclic group is abelian” is false. That is, provide a counterexample to the proposition “Every abelian group is cyclic.”
Definitions Used
- Abelian Group: A group $(G, *)$ is abelian if its binary operation is commutative, meaning $a * b = b * a$ for all $a, b \in G$.
- Cyclic Group: A group $G$ is cyclic if it can be generated by a single element. A finite group of order $n$ is cyclic if and only if it contains an element of order $n$.
- The Klein Four-Group ($V_4$): An abelian group of order 4, defined as $V_4 = \{e, a, b, c\}$, where every non-identity element has an order of 2.
Resolution
To prove the statement is false, we need to find a group that is abelian but is not cyclic. The standard counterexample is the Klein four-group, $V_4$. We must show two things about $V_4$: that it is abelian, and that it is not cyclic.
1. The Klein Four-Group is Abelian
The Klein four-group $V_4 = \{e, a, b, c\}$ is defined as an abelian group. Its operation is commutative by definition (e.g., $a*b = c$ and $b*a=c$). Therefore, it satisfies the condition of being an abelian group.
2. The Klein Four-Group is Not Cyclic
For a group to be cyclic, it must have a generator element whose order is equal to the order of the group.
- The order of the group is $|V_4| = 4$.
- To be cyclic, $V_4$ must contain an element of order 4.
- Let’s examine the orders of the elements in $V_4$:
- $\text{ord}(e) = 1$
- $\text{ord}(a) = 2$ (since $a^2=e$)
- $\text{ord}(b) = 2$ (since $b^2=e$)
- $\text{ord}(c) = 2$ (since $c^2=e$)
Since no element in the group has an order of 4, the group has no generator. Therefore, $V_4$ is not a cyclic group.
Conclusion
The Klein four-group ($V_4$) is abelian but not cyclic. This serves as a direct counterexample, proving that the statement “Every abelian group is cyclic” is false.