Question:
Let \(G\) be a finite group of order \(28.\)
Assume that \(G\) contains a subgroup of order \(7.\)
Which of the following statements is/are true?
Let \(G\) be a finite group of order \(28.\)
Assume that \(G\) contains a subgroup of order \(7.\)
Which of the following statements is/are true?
Show Hint
If a Sylow \(p\)-subgroup is unique, it is automatically normal.
This is a key property used frequently in group classification.
Updated On: Dec 6, 2025
- \(G\) contains a unique subgroup of order 7.
- \(G\) contains a normal subgroup of order 7.
- \(G\) contains no normal subgroup of order 7.
- \(G\) contains at least two subgroups of order 7.
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The Correct Option is A, B
Solution and Explanation
Step 1: Apply Sylow’s theorems.
For \(|G| = 28 = 2^2 \times 7,\) let \(n_7\) denote the number of Sylow 7-subgroups. By Sylow’s theorem, \[ n_7 \equiv 1 \ (\text{mod } 7), \quad n_7 \mid 4. \] Hence, \(n_7 = 1.\)
Step 2: Conclusion.
Since the Sylow 7-subgroup is unique, it must be normal. Therefore, (B) is correct.
For \(|G| = 28 = 2^2 \times 7,\) let \(n_7\) denote the number of Sylow 7-subgroups. By Sylow’s theorem, \[ n_7 \equiv 1 \ (\text{mod } 7), \quad n_7 \mid 4. \] Hence, \(n_7 = 1.\)
Step 2: Conclusion.
Since the Sylow 7-subgroup is unique, it must be normal. Therefore, (B) is correct.
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