Prove that the multiplicative group of integers modulo 2N can be generated by -1 and 5 when N is a power of two. | Step-by-Step Solution

TinyProf's Solution Guide 📚

1. What We're Solving:

We need to prove that when N is a power of 2, every element in Z*_{2N} (the group of integers relatively prime to 2N under multiplication mod 2N) can be expressed as some combination of powers of -1 and 5.

2. The Approach:

Think of this like showing two keys can unlock every door in a building! We need to:

• Understand what Z*_{2N} looks like when N = 2^k
• Show that -1 and 5 together can "reach" every element
• Use the structure theorem for finite abelian groups as our roadmap

The key insight is that Z*_{2N} has a specific structure when N is a power of 2, and we can exploit this!

3. Step-by-Step Solution:

Step 1: Determine the structure of Z*_{2N} When N = 2^k (k ≥ 1), we have 2N = 2^{k+1}.

• The elements of Z*_{2N} are all odd integers from 1 to 2N-1
• So |Z*_{2N}| = φ(2N) = φ(2^{k+1}) = 2^{k+1} - 2^k = 2^k = N

Step 2: Understand the group structure For 2N = 2^{k+1} where k ≥ 2, we have: Z*_{2N} ≅ Z_2 × Z_{2^{k-1}}

Step 3: Analyze the element -1

• Notice that -1 ≡ 2N-1 (mod 2N)
• (-1)² ≡ 1 (mod 2N)
• So -1 generates a subgroup of order 2, which corresponds to the Z_2 factor

Step 4: Analyze the element 5 We need to show:

• 5 has order 2^{k-1} in Z*_{2N}
• The subgroup generated by 5 intersects trivially with ⟨-1⟩

Step 5: Verify 5 has the right order

• We can show by induction that 5^{2^{k-2}} ≡ 2^k + 1 (mod 2^{k+1})
• This means 5^{2^{k-1}} ≡ 1 (mod 2^{k+1}) but 5^{2^{k-2}} ≢ 1 (mod 2^{k+1})
• Therefore, ord(5) = 2^{k-1}

Step 6: Show the generators work together Since:

• ⟨-1⟩ has order 2
• ⟨5⟩ has order 2^{k-1}
• These orders are coprime, so ⟨-1⟩ ∩ ⟨5⟩ = {1}
• |⟨-1, 5⟩| = |⟨-1⟩| × |⟨5⟩| = 2 × 2^{k-1} = 2^k = |Z*_{2N}|

4. The Answer:

Since ⟨-1, 5⟩ is a subgroup of Z_{2N} with the same order as Z_{2N}, we conclude that ⟨-1, 5⟩ = Z_{2N}. Therefore, -1 and 5 generate Z_{2N} when N is a power of 2! 🎉

5. Memory Tip:

Remember "-1 flips, 5 cycles"! The element -1 handles the "sign flipping" (Z_2 part), while 5 cycles through the main structure (Z_{2^{k-1}} part). Together, they cover everything!

Pro tip: The choice of 5 isn't random - it's the smallest number that works for all cases. Try working through small examples like N = 4, 8 to see the pattern in action!