Question:
The number of subgroups of \(Z_{48}\) is
The number of subgroups of \(Z_{48}\) is
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A cyclic group of order \(n\) has exactly one subgroup for each divisor of \(n\).
Updated On: Jul 3, 2026
- \(10\)
- \(48\)
- \(2\)
- \(8\)
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The Correct Option is A
Solution and Explanation
Concept:
For a cyclic group \(Z_n\), there is exactly one subgroup corresponding to each positive divisor of \(n\). Therefore, the number of subgroups of \(Z_n\) is equal to the number of positive divisors of \(n\).
Step 1: Identify the group. The given group is \[ Z_{48} \] This is a cyclic group of order \[ 48 \]
Step 2: Prime factorize \(48\). \[ 48=16\times 3 \] \[ 48=2^4\times 3^1 \]
Step 3: Find the number of divisors. If \[ n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k} \] then the number of positive divisors is \[ (a_1+1)(a_2+1)\cdots(a_k+1) \] Here, \[ 48=2^4\times 3^1 \] So the number of divisors is \[ (4+1)(1+1) \] \[ =5\times 2 \] \[ =10 \]
Step 4: Final conclusion. Since \(Z_{48}\) is cyclic, the number of subgroups is equal to the number of divisors of \(48\). Therefore, \[ \boxed{10} \]
For a cyclic group \(Z_n\), there is exactly one subgroup corresponding to each positive divisor of \(n\). Therefore, the number of subgroups of \(Z_n\) is equal to the number of positive divisors of \(n\).
Step 1: Identify the group. The given group is \[ Z_{48} \] This is a cyclic group of order \[ 48 \]
Step 2: Prime factorize \(48\). \[ 48=16\times 3 \] \[ 48=2^4\times 3^1 \]
Step 3: Find the number of divisors. If \[ n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k} \] then the number of positive divisors is \[ (a_1+1)(a_2+1)\cdots(a_k+1) \] Here, \[ 48=2^4\times 3^1 \] So the number of divisors is \[ (4+1)(1+1) \] \[ =5\times 2 \] \[ =10 \]
Step 4: Final conclusion. Since \(Z_{48}\) is cyclic, the number of subgroups is equal to the number of divisors of \(48\). Therefore, \[ \boxed{10} \]
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