# Verifying a Proof about Maximal Subgroups of Cyclic Groups

1. Jan 9, 2017

### Bashyboy

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
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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. Jan 9, 2017

### 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.