Question:
Let \(H=\{1,15\}\) is a subgroup of a group of order \(8\). Then the number of all left cosets of \(H\) in \(G\) is
Let \(H=\{1,15\}\) is a subgroup of a group of order \(8\). Then the number of all left cosets of \(H\) in \(G\) is
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Number of left cosets of \(H\) in \(G\) is \(\dfrac{|G|}{|H|}\).
Updated On: Jul 3, 2026
- \(2\)
- \(3\)
- \(4\)
- \(8\)
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The Correct Option is C
Solution and Explanation
Concept:
The number of left cosets of a subgroup \(H\) in a group \(G\) is called the index of \(H\) in \(G\). It is given by \[ [G:H]=\frac{O(G)}{O(H)} \]
Step 1: Find the order of \(H\). Given, \[ H=\{1,15\} \] So the number of elements in \(H\) is \[ O(H)=2 \]
Step 2: Write the order of \(G\). The group \(G\) has order \[ O(G)=8 \]
Step 3: Find the number of left cosets. \[ [G:H]=\frac{O(G)}{O(H)} \] \[ [G:H]=\frac{8}{2} \] \[ [G:H]=4 \]
Step 4: Final answer. \[ \boxed{4} \]
The number of left cosets of a subgroup \(H\) in a group \(G\) is called the index of \(H\) in \(G\). It is given by \[ [G:H]=\frac{O(G)}{O(H)} \]
Step 1: Find the order of \(H\). Given, \[ H=\{1,15\} \] So the number of elements in \(H\) is \[ O(H)=2 \]
Step 2: Write the order of \(G\). The group \(G\) has order \[ O(G)=8 \]
Step 3: Find the number of left cosets. \[ [G:H]=\frac{O(G)}{O(H)} \] \[ [G:H]=\frac{8}{2} \] \[ [G:H]=4 \]
Step 4: Final answer. \[ \boxed{4} \]
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