Question

Let G be finite set, which is a group under some operation and H be a subgroup and $a \in H$. Considering their order cardinality in increasing order :
A. $O(G)$

B. $O(a)$

C. $O(H)$

D. $Card(2^G)$

Choose the correct answer from the options given below :

• A. B, C, A, D
• B. D, B, C, A
• C. D, B, A, C
• D. B, C, D, A

Hint:

In group theory hierarchy: Element $\leq$ Subgroup $\leq$ Group. Power sets always result in the largest cardinality due to exponential growth.

Answer 1

The Correct Option is A

Concept:
• Group Theory properties define the relationships between orders of elements, subgroups, and parent groups.
• Lagrange's Theorem: For any finite group \(G\), the order of every subgroup \(H\) of \(G\) divides the order of \(G\).
• The order of an element \(a\) is the order of the cyclic subgroup generated by \(a\).

Step 1: Relate the element order and subgroup order
Since \(a \in H\), the cyclic subgroup \(\langle a \rangle\) generated by \(a\) is a subgroup of \(H\).
By Lagrange's theorem applied to \(H\): \(O(a) \text{ divides } O(H)\).
Therefore, \(O(a) \le O(H)\).

Step 2: Relate the subgroup order and group order
Since \(H\) is a subgroup of \(G\), by Lagrange's theorem applied to \(G\): \(O(H) \text{ divides } O(G)\).
Therefore, \(O(H) \le O(G)\).

Step 3: Evaluate the cardinality of the power set
\(2^G\) represents the power set of \(G\).
The cardinality is given by \(Card(2^G) = 2^{O(G)}\).
For any finite set of size \(n > 0\), \(2^n > n\). Thus, \(O(G) < Card(2^G)\).

Step 4: Arrange the sequence
The strictly increasing order is \(O(a) \le O(H) \le O(G) < Card(2^G)\).
Labels: B \(\rightarrow\) C \(\rightarrow\) A \(\rightarrow\) D.