|A/pA|<=p, finite index=> isomorphism - OIMU 2008 Problem 7

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|A/pA|<=p, finite index=> isomorphism - OIMU 2008 Problem 7

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#1 h Aug 28, 2010, 12:42 AM • 2 Y

Y by Adventure10, PikaPika999

Let $A$ be an abelian additive group such that all nonzero elements have infinite order and for each prime number $p$ we have the inequality $|A/pA|\leq p$ , where $pA = \{pa |a \in A\}$ , $pa = a+a+\cdots+a$ (where the sum has $p$ summands) and $|A/pA|$ is the order of the quotient group $A/pA$ (the index of the subgroup $pA$ ).

Prove that each subgroup of $A$ of finite index is isomorphic to $A$ .

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#2 h Aug 28, 2010, 1:27 AM • 3 Y

Y by Adventure10, Mango247, PikaPika999

If $[A:B]=n$ , show that $nA\subset B$ . Now use the hypotheses to show $[A:nA]\le n$ .

Steve

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pi_quadrat_sechstel

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#3 h May 15, 2025, 8:00 PM • 1 Y

Y by PikaPika999

Jorge Miranda wrote:

Let $A$ be an abelian additive group such that all nonzero elements have infinite order and for each prime number $p$ we have the inequality $|A/pA|\leq p$ , where $pA = \{pa |a \in A\}$ , $pa = a+a+\cdots+a$ (where the sum has $p$ summands) and $|A/pA|$ is the order of the quotient group $A/pA$ (the index of the subgroup $pA$ ).

Prove that each subgroup of $A$ of finite index is isomorphic to $A$ .

Claim 1: For every finite index subgroup $B$ of $A$ , the group $A/B$ is cyclic.

Proof: $B/A$ is a finite abelian group, so it is isomorphic to a group of the form $\mathbb{Z}/p_1^{e_1}\mathbb{Z}\times\cdots\times\mathbb{Z}/p_k^{e_k}\mathbb{Z}$ . No prime $p$ can appear twice in this product. (Otherwise $|A/pA|$ would not be $\leq p$ .) But for pairwise diffrent $p_1,\ldots,p_k$ the group $\mathbb{Z}/p_1^{e_1}\mathbb{Z}\times\cdots\times\mathbb{Z}/p_k^{e_k}\mathbb{Z}$ is cyclic.

Claim 2: The groups of the form $A/nA$ are all finite.

Proof: Since $A/p_1^{e_1}\cdots p_k^{e_k}A\cong A/p_1^{e_1}A\times\cdots\times A/p_k^{e_k}A$ it suffices to consider the case $n=p^e$ . In this group, the order of every element is one of the numbers $1,p,\ldots,p^e$ . By the finiteness of $A/pA$ we get one by one, that there are finitely many elements of $A/pA$ with order $p^e$ , finitely many elements of $A/pA$ with order $p^{e-1}$ and so on. So $A/p^eA$ is finite.

So $A/nA\cong \mathbb{Z}/d\mathbb{Z}$ for a $d\in\mathbb{Z}$ . We must have $d\mid n$ since all orders in $A/nA$ divide $n$ .

If $B$ is an index $n$ subgroup of $A$ , we have $A/B\cong\mathbb{Z}/n\mathbb{Z}$ and $nA\subseteq B$ . So we get a surjective map $A/nA\to A/B$ . But by the result above, this is only possible for $A/nA\cong\mathbb{Z}/n\mathbb{Z}$ and $B=nA$ .

Finally, we notice that the surjective group homeomorphism $A\to nA, a\mapsto na$ is also bijective by the infinite order condition of the problem statement. Thus $B\cong nA\cong A$ .

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