Proof: The Converse of “Subgroup of a Cyclic Group is Cyclic” is Not True
Problem
Show that the converse of the theorem “every subgroup of a cyclic group is cyclic” is not true. Specifically, provide a counterexample to the statement: “If every proper subgroup of a group $G$ is cyclic, then $G$ must be cyclic.”
Definitions Used
- Cyclic Group: A group $G$ is cyclic if there exists an element $g \in G$ (a generator) such that $G = \langle g \rangle = \{ g^n \mid n \in \mathbb{Z} \}$. For a finite group of order $n$, this is equivalent to having an element of order $n$.
- Proper Subgroup: A subgroup of a group $G$ that is not equal to $G$ itself.
- Order of an Element: The order of an element $g \in G$, denoted $\text{ord}(g)$, is the smallest positive integer $n$ such that $g^n = e$, where $e$ is the identity element.
- The Klein Four-Group ($V_4$): An abelian group of order 4, defined as $V_4 = \{e, a, b, c\}$. Its operation is defined such that every element is its own inverse ($g^2 = e$ for all $g \in V_4$) and the product of any two non-identity elements is the third (e.g., $ab=c$).
Resolution
To prove the statement is false, we will provide a counterexample. The group we will use is the Klein four-group, $V_4$. We must show that $V_4$ satisfies the condition (all its proper subgroups are cyclic) but does not satisfy the conclusion (it is not cyclic itself).
Part 1: All Proper Subgroups of $V_4$ are Cyclic
The order of $V_4$ is 4. By Lagrange’s Theorem, the possible orders of its subgroups are 1, 2, and 4. The proper subgroups, therefore, can only have orders 1 or 2.
- Subgroup of order 1: The only subgroup of order 1 is the trivial subgroup, $\{e\}$. This is cyclic, as it is generated by the identity element: $\langle e \rangle = \{e\}$.
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Subgroups of order 2: Any group of order 2 is cyclic. The subgroups of order 2 in $V_4$ are:
- $H_1 = \{e, a\}$, which is generated by $a$ since $a^1=a$ and $a^2=e$. Thus, $H_1 = \langle a \rangle$.
- $H_2 = \{e, b\}$, which is generated by $b$ since $b^1=b$ and $b^2=e$. Thus, $H_2 = \langle b \rangle$.
- $H_3 = \{e, c\}$, which is generated by $c$ since $c^1=c$ and $c^2=e$. Thus, $H_3 = \langle c \rangle$.
We have examined all possible proper subgroups and confirmed that every one is cyclic.
Part 2: The Group $V_4$ Itself is Not Cyclic
For a group to be cyclic, it must contain a generator, which is an element whose order is equal to the order of the group.
- The order of the group is $|V_4| = 4$.
- The orders of its elements are:
- $\text{ord}(e) = 1$
- $\text{ord}(a) = 2$ (since $a^2 = e$)
- $\text{ord}(b) = 2$ (since $b^2 = e$)
- $\text{ord}(c) = 2$ (since $c^2 = e$)
Since no element in $V_4$ has an order of 4, the group does not have a generator. Therefore, $V_4$ is not a cyclic group.
Conclusion
The Klein four-group, $V_4$, is a non-cyclic group where every proper subgroup is cyclic. This serves as a direct counterexample to the statement “If every proper subgroup of a group $G$ is cyclic, then $G$ must be cyclic.”
Thus, the converse is not true.