# Thread: subgroups of a factor group

1. ## subgroups of a factor group

Hi! I have a question concerning subgroups: Define $\displaystyle H_p=\{\frac{n}{p^\alpha}: n, \alpha \in \mathbb{Z}, \alpha \geq 0\}$, where p is a prime. Then $\displaystyle H_p$ is a subgroup of $\displaystyle (\mathbb{Q},+)$. I want to show:

(1) In the quotient group $\displaystyle \mathbb{Q} / \mathbb{Z}$ the subgroup $\displaystyle G_p=H_p / \mathbb{Z}$ is exactly the set of elements with order a power of p (EDIT).

(2) What are the real subgroups of $\displaystyle G_p$?

Does anyone have ideas for how to solve this? I am thankful for any hints.
Banach

2. Originally Posted by Banach
Hi! I have a question concerning subgroups: Define $\displaystyle H_p=\{\frac{n}{p^\alpha}: n, \alpha \in \mathbb{Z}, \alpha \geq 0\}$, where p is a prime. Then $\displaystyle H_p$ is a subgroup of $\displaystyle (\mathbb{Q},+)$. I want to show:

(1) In the quotient group $\displaystyle \mathbb{Q} / \mathbb{Z}$ the subgroup $\displaystyle G_p=H_p / \mathbb{Z}$ is exactly the set of elements with order p.
i haven't thought about yet but this part doesn't seem quite right to me, because $\displaystyle \frac{1}{p^2} + \mathbb{Z} \in G_p$ is not of order $\displaystyle p.$

3. Oh yes, thank you. It should be the set of elemtents with order a power of p. I will change it.

4. Originally Posted by Banach
Hi! I have a question concerning subgroups: Define $\displaystyle H_p=\{\frac{n}{p^\alpha}: n, \alpha \in \mathbb{Z}, \alpha \geq 0\}$, where p is a prime. Then $\displaystyle H_p$ is a subgroup of $\displaystyle (\mathbb{Q},+)$. I want to show:

(1) In the quotient group $\displaystyle \mathbb{Q} / \mathbb{Z}$ the subgroup $\displaystyle G_p=H_p / \mathbb{Z}$ is exactly the set of elements with order a power of p (EDIT).
let $\displaystyle r = \frac{a}{b} \in \mathbb{Q}, \ \gcd(a,b)=1.$ then the order of $\displaystyle r + \mathbb{Z}$ is a power of p if and only if $\displaystyle p^k r =\frac{p^ka}{b} \in \mathbb{Z},$ for some integer $\displaystyle k \geq 0$, which is equivalent to say that $\displaystyle b \mid p^k$, since $\displaystyle \gcd(a,b)=1.$

thus $\displaystyle b=p^n,$ for some $\displaystyle 0 \leq n \leq k.$ hence $\displaystyle r=\frac{a}{p^n} \in H_p.$

(2) What are the real subgroups of $\displaystyle G_p$?
the subgroups of $\displaystyle G_p$ are exactly in the form $\displaystyle N/\mathbb{Z},$ where $\displaystyle N$ is a subgroup of $\displaystyle H_p,$ which (properly) contains $\displaystyle \mathbb{Z}.$ since we're looking for proper subgroups, we have $\displaystyle N \neq H_p.$ we need a lemma first:

Lemma: if $\displaystyle \frac{a}{p^k} \in N,$ for some $\displaystyle k \geq 0$ and $\displaystyle a \in \mathbb{Z}$ with $\displaystyle \gcd(a,p)=1,$ then $\displaystyle \frac{n}{p^s} \in N,$ for all $\displaystyle n \in \mathbb{Z}, \ 0 \leq s \leq k.$

Proof: since $\displaystyle \gcd(a,p^k)=1,$ there exist integers $\displaystyle u,v$ such that $\displaystyle up^k+va=1.$ thus: $\displaystyle \frac{1}{p^k}=u+v\frac{a}{p^k} \in N.$ hence, if $\displaystyle n \in \mathbb{Z}$ and $\displaystyle 0 \leq s \leq k,$ then $\displaystyle \frac{n}{p^s}=np^{k-s}\frac{1}{p^k} \in N. \ \ \Box$

now let $\displaystyle A=\{k \geq 0: \ \frac{a}{p^k} \in N, \ \text{for some} \ a \in \mathbb{Z} \ \text{with} \ \gcd(a,p)=1 \}.$ if $\displaystyle A$ is unbounded, then by the Lemma, we will have $\displaystyle N=H_p,$ which is not a proper subgroup. otherwise, let $\displaystyle m=\max A.$

then again the Lemma gives us: $\displaystyle N=\{\frac{n}{p^s}: \ n \in \mathbb{Z}, \ 0 \leq s \leq m \}=\frac{1}{p^m}\mathbb{Z}.$ since $\displaystyle \mathbb{Z} \subsetneq N,$ we must have $\displaystyle m \geq 1.$ so the set $\displaystyle \left \{\frac{\frac{1}{p^m}\mathbb{Z}}{\mathbb{Z}}: \ m \geq 1 \right \}$ is exactly the set of all proper subgroups of $\displaystyle G_p. \ \ \Box$

5. Thank you very, very much for your so well-structured answer. My problem was that i always thought of the wrong group operation, namely $\displaystyle (\mathbb{Q},\cdot)$.

Nice greetings
Banach