## Review of Group Theory: Cyclic Groups and Cyclic Subgroups (Pt. II)

**Point of post: **This post is a continuation of this one.

Our last theorem we wish to prove is the most important of all. Indeed, it will completely classify all subgroups of cyclic groups. Unsurprisingly this theorem is called the *fundamental theorem of cyclic groups.*

**Theorem: ***Let . Then:*

**Proof: **

: To do this let . If this is trivial, so assume not and let . We claim that . To do this note that for any we have( by the division algorithm) that with and . Note though that since and is a subgroup we have that and thus . But, if we would have that and , but this contradicts the minimality of . It follows that . Thus, since was arbitrary it follows that and the opposite containment follows from closure of . Thus, as required.

We first claim that if are distinct then . But, this is clear. If or is zero this is trivial, so assume that . Then if then we must have that and so there exists such that or that but this is true if and only if and so it clearly follows that . Applying the exact same argument is reverse shows that and by assumption this can only be true if . But, by and noticing that (clearly) we know that these are the only such subgroups from where the conclusion follows.

: Let . There is clearly at least one subgroup of order since and

To see that this is the only such subgroup suppose that and . Note then that

comparing the two we find that and thus . It clearly follows then that every multiple of is a multiple of and so

but since the two sets have the same cardinality it follows that as desired.

Thus, it remains to show that if (we clearly don’t need to consider the trivial subgroup, since the proof follows immediately) and then . To do this note that by we have that where . Note though that

But, from the division algorithm we know that there exists with such that . We see then that

and since and it follows that so that as desired. The conclusion follows

**References:**

1. Lang, Serge. *Undergraduate Algebra*. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

[…] every cyclic group has precisely one subgroup of each order which divides the ambient group (see here for a proof). So, assume that is not cyclic and let be such that . If we’re done since […]

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[…] order is invariant under conjugation and so in particular but since a cyclic group has precisely one subgroup of every order dividing it we may conclude that and so in particular . The conclusion […]

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