- #1

Bashyboy

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- 5

## Homework Statement

Show that if ##G = \langle x \rangle## is a cyclic group of order ##n \ge 1##, then a subgroup ##H## is maximal; if and only if ##H = \langle x^p \rangle## for some prime ##p## dividing ##n##

## Homework Equations

A subgroup ##H## is called maximal if ##H \neq G## and the only subgroups of ##G## which contain ##H## are itself and ##G##.

## The Attempt at a Solution

First we prove that if ##H = \langle x^p \rangle## for some prime ##p## dividing ##n##, then necessarily ##H## is maximal in ##G##. Assume that ##K < G## is some subgroup containing ##H##. Since every subgroup of a cyclic group is cyclic, we see that ##K = \langle x^k \rangle##, where ##k## is the smallest positive integer for which ##x^k \in K##. Furthermore, ##\langle x^p \rangle \le \langle x^k \rangle## implies that ##\frac{|x^k|}{|x^p|} = \frac{n/(n,k)}{n/(n,p)} = \frac{p}{(n,k)}##, where we used ##(n,p) = p## because ##p## is a prime dividing ##n##, and this implies either ##(n,k)=1## or ##(n,k)=p##. In the former case, ##K = G##, while in the latter case ##K = H##. In either case, we have that there is no subgroup distinct from either ##H## or ##G## containing ##H##, indicating that ##H## is maximal in ##G##

Now we prove that if ##H## is maximal, then ##H = \langle x^p \rangle## for some prime ##p## dividing ##n##. Since ##H## is a subgroup the of the cyclic group ##G##, then ##H = \langle x^p \rangle##, where ##p## is the smallest natural number for which ##x^p \in H##. By way of contradiction, suppose that ##p## is composite, i.e., ##p = ab## for positive coprime integers ##a## and ##b##. Then ##H## is a subgroup of both ##\langle x^b \rangle## and ##\langle x^b \rangle##. Note that ##\langle x^a \rangle = G = \langle x \rangle## cannot be true, since this would entail ##(a,n)=1##, which is not true because ##a## divides ##ab = p## and ##p## divides ##n##. Likewise, ##\langle x^b \rangle \neq G##. The only remaining possibility is ##\langle x^a \rangle = H = \langle x^b \rangle##. But this implies ##|x^a| = |x^b|## or ##\frac{n}{(a,n)} = \frac{n}{(b,n)}## or ##(a,n) = (b,n)##, contradicting the fact that ##a## and ##b## are relatively prime. Thus, ##p## must be prime dividing ##n##, since ##\langle x^p = \rangle = \langle x^{(p,n)} \rangle##.

I welcome criticisms and suggestions for a shorter proof, if possible.