About this video

• Video Title: Every subgroup of a cyclic group is cyclic.
• Channel: Maths ICU
• Speakers: None
• Duration: 00:08:17

Overview

This video presents and proves the theorem stating that every subgroup of a cyclic group is also cyclic. The proof involves assuming a cyclic group G generated by an element A, taking an arbitrary subgroup H of G, and demonstrating that H is also cyclic by identifying a generator for it.

Key takeaways

• Theorem Statement: The core theorem asserts that if a group is cyclic, then any subgroup within it must also be cyclic.
• Proof Setup: The proof begins by defining a cyclic group G generated by an element A, and then considers an arbitrary subgroup H of G.
• Identifying a Generator: The proof focuses on finding a generator for the subgroup H. It defines 'k' as the smallest positive integer such that A raised to the power of k is an element of H.
• Using Division Algorithm: The proof employs the division algorithm to show that if an element B in H can be expressed as A to the power of M, then M must be a multiple of k. This is achieved by assuming M is not a multiple of k, leading to a contradiction.
• Conclusion: By demonstrating that every element in H can be expressed as a power of A raised to the power of k, it is concluded that A raised to the power of k is the generator of H, thus proving H is cyclic.

The contradiction arises in the proof when it's assumed that 'k' (the smallest positive integer such that A^k is in H) does not divide 'm' (where B = A^m is an arbitrary element in H).

Here's a breakdown:

1. The Setup: We have a cyclic group G generated by A, and a subgroup H. We defined 'k' as the smallest positive integer such that A^k is in H. We then take any element B in H, and since G is cyclic and generated by A, B can be written as A^m for some integer m.

2. The Assumption (for contradiction): We assume that 'k' does not divide 'm'.

3. Applying the Division Algorithm: If 'k' does not divide 'm', the division algorithm states that we can write: m = qk + r where 'q' is the quotient and 'r' is the remainder. Crucially, for this to be a non-division scenario, the remainder 'r' must be greater than 0 (r > 0) and strictly less than 'k' (r < k).

4. Manipulating the Expression: The video then considers A^r. By substituting the expression for 'm': A^r = A^(m - qk) Using exponent rules, this can be rewritten as: A^r = A^m * A^(-qk) Since B = A^m and A^qk = (A^k)^q, we get: A^r = B * (A^k)^(-q)

5. The Crucial Insight:

   • We know B is in H (our arbitrary element).
   • We know A^k is in H (by definition of k).
   • Since H is a subgroup, it's closed under its operations. This means if B is in H and A^k is in H, then any combination of them, including (A^k)^(-q) and their product B * (A^k)^(-q), must also be in H.
   • Therefore, A^r must be in H.

6. The Contradiction:

   • We found that A^r is in H.
   • We also know that 0 < r < k.
   • This means we have found a positive integer 'r' which is smaller than 'k', and A^r is an element of H.
   • However, we initially defined 'k' as the smallest positive integer for which A^k is in H.
   • Finding such an 'r' contradicts the definition of 'k'.

This contradiction forces us to reject our initial assumption that 'k' does not divide 'm'. Therefore, 'k' must divide 'm'. The rest of the proof follows from this established fact.