Proving a Group is Cyclic - A Guide for Beginners

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SUMMARY

To prove that a group K of order 143 is cyclic, it is essential to demonstrate the existence of normal subgroups of order 11 and 13. Utilizing Sylow's theorems, one can establish that there is exactly one normal subgroup of each order in K. Since the only groups of these orders are Z_11 and Z_13, both of which are cyclic, their direct product is also cyclic. This approach is applicable for prime pairs where neither prime is congruent to 1 modulo the other.

PREREQUISITES
  • Understanding of group theory concepts, specifically cyclic groups
  • Familiarity with Sylow's theorems
  • Knowledge of normal subgroups and their properties
  • Basic comprehension of direct products of groups
NEXT STEPS
  • Study Sylow's theorems in detail to understand their implications on group structure
  • Learn about the properties of normal subgroups and their significance in group theory
  • Explore examples of cyclic groups and their direct products
  • Investigate groups of composite order and their classifications
USEFUL FOR

Mathematicians, students of abstract algebra, and anyone interested in group theory, particularly those looking to deepen their understanding of cyclic groups and their properties.

friedchicken
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Hi everyone.

How could I prove if something is a cyclic group? I was wondering because I can prove is something is a group, a subgroup, and a normal subgroup, but I have no Idea as to how to prove something is a cyclic group.

Ex: Suppose K is a group with order 143. Prove K is cyclic.

I read around and I kept seeing something about Sylow's theorems, but I never learned anything like that. Is there another approach?
 
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To prove that K is cyclic, it is sufficient to prove that it has normal subgroups of order 11 and 13. There's a theorem that states that, if we can find such normal subgroups and their intersection is trivial, then K is isomorphic to their direct product. The only groups of order 11 and 13 are Z_11 and Z_13. Since both are cyclic and 11 and 13 are coprime, their direct product is cyclic.

By Sylow's theorems, there's exactly one of each in K.

This works quite generally for prime pairs as long as p\not = 1 mod q and q\not =1 mod p. If p = 1 mod q (such would be the case, for example, if one of the primes is 2), the subgroup of order 2 is not unique and therefore not normal: therefore, for example, there's a non-cyclic group of order 6, S_3 (3 = 1 mod 2), and there's a nontrivial non-cyclic group of order 21 (7 = 1 mod 3).

The same method goes if there are more than two primes in the decomposition.
 
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