# subgroups of a factor group

• Oct 21st 2008, 11:34 AM
Banach
subgroups of a factor group
Hi! I have a question concerning subgroups: Define $H_p=\{\frac{n}{p^\alpha}: n, \alpha \in \mathbb{Z}, \alpha \geq 0\}$, where p is a prime. Then $H_p$ is a subgroup of $(\mathbb{Q},+)$. I want to show:

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

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

Does anyone have ideas for how to solve this? I am thankful for any hints.
Banach
• Oct 21st 2008, 11:49 AM
NonCommAlg
Quote:

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

(1) In the quotient group $\mathbb{Q} / \mathbb{Z}$ the subgroup $G_p=H_p / \mathbb{Z}$ 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 $\frac{1}{p^2} + \mathbb{Z} \in G_p$ is not of order $p.$
• Oct 21st 2008, 12:08 PM
Banach
Oh yes, thank you. It should be the set of elemtents with order a power of p. I will change it.
• Oct 22nd 2008, 01:07 AM
NonCommAlg
Quote:

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

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

let $r = \frac{a}{b} \in \mathbb{Q}, \ \gcd(a,b)=1.$ then the order of $r + \mathbb{Z}$ is a power of p if and only if $p^k r =\frac{p^ka}{b} \in \mathbb{Z},$ for some integer $k \geq 0$, which is equivalent to say that $b \mid p^k$, since $\gcd(a,b)=1.$

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

Quote:

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

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

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

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

then again the Lemma gives us: $N=\{\frac{n}{p^s}: \ n \in \mathbb{Z}, \ 0 \leq s \leq m \}=\frac{1}{p^m}\mathbb{Z}.$ since $\mathbb{Z} \subsetneq N,$ we must have $m \geq 1.$ so the set $\left \{\frac{\frac{1}{p^m}\mathbb{Z}}{\mathbb{Z}}: \ m \geq 1 \right \}$ is exactly the set of all proper subgroups of $G_p. \ \ \Box$
• Oct 22nd 2008, 08:41 AM
Banach
Thank you very, very much for your so well-structured answer. My problem was that i always thought of the wrong group operation, namely $(\mathbb{Q},\cdot)$.

Nice greetings
Banach