Question:
Let \(G\) be a finite group. Then \(G\) is necessarily a cyclic group if the order of \(G\) is
Let \(G\) be a finite group. Then \(G\) is necessarily a cyclic group if the order of \(G\) is
Show Hint
Every group of prime order is cyclic.
Updated On: Jul 3, 2026
- \(4\)
- \(7\)
- \(6\)
- \(10\)
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The Correct Option is B
Solution and Explanation
Concept:
Every group of prime order is cyclic. If \[ O(G)=p \] where \(p\) is prime, then any non-identity element of \(G\) generates the entire group.
Step 1: Check the given orders. The options are \[ 4,\quad 7,\quad 6,\quad 10 \] Among these, only \[ 7 \] is prime.
Step 2: Use the theorem. If a group has prime order \(p\), then by Lagrange's theorem, the order of any non-identity element must divide \(p\). The divisors of \(p\) are \[ 1\quad \text{and}\quad p \] A non-identity element cannot have order \(1\), so it must have order \(p\). Thus it generates the whole group.
Step 3: Apply to \(O(G)=7\). Since \(7\) is prime, any group of order \(7\) is cyclic.
Step 4: Final answer. \[ \boxed{7} \]
Every group of prime order is cyclic. If \[ O(G)=p \] where \(p\) is prime, then any non-identity element of \(G\) generates the entire group.
Step 1: Check the given orders. The options are \[ 4,\quad 7,\quad 6,\quad 10 \] Among these, only \[ 7 \] is prime.
Step 2: Use the theorem. If a group has prime order \(p\), then by Lagrange's theorem, the order of any non-identity element must divide \(p\). The divisors of \(p\) are \[ 1\quad \text{and}\quad p \] A non-identity element cannot have order \(1\), so it must have order \(p\). Thus it generates the whole group.
Step 3: Apply to \(O(G)=7\). Since \(7\) is prime, any group of order \(7\) is cyclic.
Step 4: Final answer. \[ \boxed{7} \]
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