Let 𝐺 be a group of order 52 = 4 ⋅ 13 and Let 𝐺 = ℤ/2ℤ × ℤ/26ℤ. (The operation being addition.) How many subgroups of order 4 does 𝐺 have? show




To determine the number of subgroups of order 4 in the group \( G = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/26\mathbb{Z} \), we need to explore the structure of \( G \) and identify which subsets form subgroups of the desired order.

1. **Structure of \( G \)**:
   - \( \mathbb{Z}/2\mathbb{Z} \) has elements \( \{0, 1\} \).
   - \( \mathbb{Z}/26\mathbb{Z} \) has elements \( \{0, 1, 2, \ldots, 25\} \).

   Hence, \( G \) consists of pairs \( (a, b) \) where \( a \in \mathbb{Z}/2\mathbb{Z} \) and \( b \in \mathbb{Z}/26\mathbb{Z} \).

2. **Order of elements**:
   - The order of an element \( (a, b) \) in \( G \) is the least common multiple (LCM) of the orders of \( a \) and \( b \).
   - The order of \( a \in \mathbb{Z}/2\mathbb{Z} \) is 1 if \( a = 0 \) and 2 if \( a = 1 \).
   - The order of \( b \in \mathbb{Z}/26\mathbb{Z} \) depends on the value of \( b \). Specifically, the order of \( b \) is the smallest positive integer \( k \) such that \( kb \equiv 0 \pmod{26} \).

3. **Finding elements of order 4**:
   - To have a subgroup of order 4, we need an element whose order is 4. This requires that \( \text{LCM}(\text{order of } a, \text{order of } b) = 4 \).

   - **Case 1**: \( a = 0 \).
     - Here, the order of \( (0, b) \) is just the order of \( b \). We need \( b \) to have order 4.
     - An element \( b \) has order 4 in \( \mathbb{Z}/26\mathbb{Z} \) if \( b \) is such that \( 26/\gcd(26, b) = 4 \). Thus, \( \gcd(26, b) = 26/4 = 6.5 \), which is not possible. So, no elements \( b \) in \( \mathbb{Z}/26\mathbb{Z} \) have order 4.

   - **Case 2**: \( a = 1 \).
     - Here, the order of \( (1, b) \) is \( \text{LCM}(2, \text{order of } b) \). To achieve order 4, the order of \( b \) must be 4.
     - The order of \( b \) in \( \mathbb{Z}/26\mathbb{Z} \) must be 4. As shown above, there are no elements in \( \mathbb{Z}/26\mathbb{Z} \) with order 4.

Since there are no elements of order 4 in \( \mathbb{Z}/26\mathbb{Z} \), there can be no elements in \( G \) with order 4, and thus no subgroups of order 4 in \( G \).

4. **Double-check**:
   - The elements in \( G \) form

Explore how to determine the number of subgroups of order 4 in the group G = Z/2Z × Z/26Z using group theory concepts. Understand the structure of G, the order of its elements, and the conditions for forming subgroups of specific orders. Suitable for advanced undergraduate students studying group theory.

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