Verifying a Proof about Maximal Subgroups of Cyclic Groups

In summary, if a subgroup H is maximal in a cyclic group G of order n, then H is equal to <x^p> for some prime p dividing n, and if H is equal to <x^p> for some prime p dividing n, then H is maximal in G. This can be proven by showing that if p is composite, then H is a subgroup of both <x^a> and <x^b>, which leads to a contradiction, ultimately proving that p must be prime.
  • #1
Bashyboy
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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.
 
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  • #2
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.
 

Related to Verifying a Proof about Maximal Subgroups of Cyclic Groups

1. What is the definition of a cyclic group?

A cyclic group is a type of group in abstract algebra that is generated by a single element. This means that all elements in the group can be written as powers of the generator element.

2. How do you prove that a subgroup of a cyclic group is maximal?

To prove that a subgroup of a cyclic group is maximal, you need to show that it is not a proper subgroup of any other subgroup. This can be done by showing that the subgroup contains all elements that are not in the original cyclic group, and that it is not possible to add any other elements to the subgroup without creating a larger subgroup.

3. What is the significance of verifying a proof about maximal subgroups of cyclic groups?

Verifying a proof about maximal subgroups of cyclic groups is important because it allows us to understand the structure and properties of cyclic groups better. It also helps us to identify and classify different types of subgroups within cyclic groups, which can be useful in various areas of mathematics and science.

4. Can a cyclic group have more than one maximal subgroup?

Yes, a cyclic group can have multiple maximal subgroups. This is because a maximal subgroup is defined as a subgroup that is not a proper subgroup of any other subgroup, but there can be multiple subgroups that meet this criteria within a cyclic group.

5. How can we use maximal subgroups of cyclic groups in real-life applications?

Maximal subgroups of cyclic groups can be used in cryptography, coding theory, and other areas of computer science. In these applications, cyclic groups and their subgroups are used to create secure and efficient algorithms for data encryption and communication.

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