# 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 $\displaystyle G$ of order $\displaystyle q-1$ where $\displaystyle q$ is prime.

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

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

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

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 $\displaystyle G$ of order $\displaystyle q-1$ where $\displaystyle q$ is prime. i guess you meant "group"!

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

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

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

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

something is not right here! any finite cyclic group of order $\displaystyle n$ has a unique (cyclic) subgroup of any order which divides $\displaystyle n.$ 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 $\displaystyle n$ has a unique (cyclic) subgroup of any order which divides $\displaystyle n.$

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 $\displaystyle \mathbb{Z}_p$ to $\displaystyle Aut(\mathbb{Z}_q)$ where p divides q-1.)

I just got stuck on proving $\displaystyle Aut(\mathbb{Z}_q)$ has a unique subgroup of order $\displaystyle p$.
• 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 $\displaystyle G=<a>$ be a group of order $\displaystyle n$ and $\displaystyle d \mid n.$ then clearly $\displaystyle <a^{\frac{n}{d}}>$ is a subgroup of $\displaystyle G$ of order $\displaystyle d.$

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

therefore, since $\displaystyle |H|=|<a^{\frac{n}{d}}>|,$ we must have $\displaystyle H=<a^{\frac{n}{d}}>.$
• Apr 28th 2009, 07:29 PM
ziggychick
ok got it. thanks