# cyclic subgroup

• Apr 28th 2009, 06:12 PM
ziggychick
cyclic subgroup
I should be able to do this one but I'm stuck...

I have a cyclic subgroup $G$ of order $q-1$ where $q$ is prime.

Let $p$ be a prime dividing $q-1$. Show G contains a unique cyclic subgroup of order $p$.

Now I know Cauchy's Theorem tells me G indeed has at least one subgroup of order $p$.

Also I know $G\simeq\mathbb{Z}_{pm}$ for some positive integer $m$.

I may be making this too complicated...Any hints?
• Apr 28th 2009, 06:50 PM
NonCommAlg
Quote:

Originally Posted by ziggychick
I should be able to do this one but I'm stuck...

I have a cyclic subgroup $G$ of order $q-1$ where $q$ is prime. i guess you meant "group"!

Let $p$ be a prime dividing $q-1$. Show G contains a unique cyclic subgroup of order $p$.

Now I know Cauchy's Theorem tells me G indeed has at least one subgroup of order $p$.

Also I know $G\simeq\mathbb{Z}_{pm}$ for some positive integer $m$.

I may be making this too complicated...Any hints?

something is not right here! any finite cyclic group of order $n$ has a unique (cyclic) subgroup of any order which divides $n.$ so, in your problem, we don't need to assume that p, q are prime .

are you sure G is cyclic?
• Apr 28th 2009, 07:00 PM
ziggychick
Quote:

Originally Posted by NonCommAlg
any finite cyclic group of order $n$ has a unique (cyclic) subgroup of any order which divides $n.$

Ok, i need to prove the above statement.

I was assuming p and q where prime because this is actually part of a larger proof. ( that there are p distinct homomorphisms from $\mathbb{Z}_p$ to $Aut(\mathbb{Z}_q)$ where p divides q-1.)

I just got stuck on proving $Aut(\mathbb{Z}_q)$ has a unique subgroup of order $p$.
• Apr 28th 2009, 07:21 PM
NonCommAlg
Quote:

Originally Posted by ziggychick

Ok, i need to prove the above statement.

that's not hard to prove:

first of all recall that every subgroup of a cyclic group is cyclic. now let $G=$ be a group of order $n$ and $d \mid n.$ then clearly $$ is a subgroup of $G$ of order $d.$

now suppose $H=$ is any subgroup of G of order $d.$ then $a^{kd}=1$ and thus $n \mid kd,$ i.e. $\frac{n}{d} \mid k.$ let $k=m\frac{n}{d}.$ then $a^k=a^{m \frac{n}{d}} \in $ and hence $H \subseteq .$

therefore, since $|H|=||,$ we must have $H=.$
• Apr 28th 2009, 07:29 PM
ziggychick
ok got it. thanks