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Verifying a Proof about Maximal Subgroups of Cyclic Groups

  1. Jan 9, 2017 #1
    1. The problem statement, all variables and given/known data
    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##

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

    3. 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.
     
  2. jcsd
  3. Jan 9, 2017 #2

    fresh_42

    Staff: Mentor

    Perhaps you can shorten the the second part a little bit. Assuming ##p## as composite, you can as well assume ##a \neq 1## w.l.o.g. This means ##\langle x^a \rangle \neq G## because ##a\,\vert \,p\,\vert \,n##. By maximality of ##H## you get ##H = \langle x^a \rangle = \langle x^p=(x^a)^b \rangle## and thus ##b=1## which means ##p=a## is prime.
     
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