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I am trying to show that for a given prime $p$ and a finite Abelian $p$-group $G$ that if $G$ has an unique subgroup of order $p$, then $G$ is necessarily cyclic.

I am trying to prove the statement, by arguing by contradiction. Of course, the statement would be easy to prove if one knows the Fundamental Theorem of Finitely Generated Abelian Groups, but I was trying to prove the statement without invoking this Theorem, and I am having trouble proving the statement.

I know that by Cauchy's Theorem, one can find an element $x$ in $G$ of order $p$; how does one then produce an element $y\in G\backslash\langle x\rangle$ of order $p$?

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  • $\begingroup$ You have to demand the finiteness for $G$. $\endgroup$ – Boris Novikov Aug 24 '13 at 16:51
  • $\begingroup$ I don't know if it works, but maybe induction on $n$ for groups of order $p^n$? The first case is trivial, each case after that you can mod out by the cyclic group $\langle x\rangle$, and use induction there? Just thinking out loud. $\endgroup$ – Clayton Aug 24 '13 at 16:52
  • $\begingroup$ Cauchy's theorem is hardly necessary when the hypotheses dictate a subgroup of order $p$. Also don't they further dictate there is no $y\in G\setminus\langle x\rangle$ of order $p$ by uniqueness? $\endgroup$ – anon Aug 24 '13 at 16:53
  • $\begingroup$ The goal is to arrive at a contradiction by finding a $y\in G\backslash\langle x\rangle$. $\endgroup$ – yoshi Aug 24 '13 at 17:01
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Let $G$ contain an unique cyclic subgroup of order $p^k$ and $a$ be an element of order $p^k$. Prove that it contains an unique cyclic subgroup of order $p^{k+1}$.

Let $x,y$ be elements of order $p^{k+1}$ and $x^{p^k}=y^{p^k}=a$. Then $(xy^{-1})^{p^k}=1$, so $x=ya^m=y^{mp^k+1}$. Further use the induction.

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  • $\begingroup$ May I ask you why is not restrictive to assume $y^{p^{k}}=x^{p^{k}}=a$? $\endgroup$ – Riccardo Aug 24 '13 at 18:34
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    $\begingroup$ @Ric Ped $x^{p^{k}}$ has the order $p$, so $x^{p^{k}}=a^l$. Let $ll'\equiv 1 \mod p$. Then $(x^{l'})^{p^{k}}=a$. $\endgroup$ – Boris Novikov Aug 24 '13 at 19:07
  • $\begingroup$ So we are "building" such elements starting from two elements of order $p^{k+1}$ right? In this way we are allowed to demand this hypothesis without loss of generalities, am I right? $\endgroup$ – Riccardo Aug 24 '13 at 20:23
  • $\begingroup$ @ Ric Ped : Yes, certainly. $\endgroup$ – Boris Novikov Aug 24 '13 at 20:33
  • $\begingroup$ Thank you very much for your patience :) $\endgroup$ – Riccardo Aug 24 '13 at 21:18
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I assume that $G$ is finite (sometimes the term "$p$-group" is used for infinite groups as well, but for infinite abelian $p$-groups the statement is false).

If it is undesirable to use FTFGAG, then I would proceed as follows.

  1. Introduce the endomorphism $\varphi: G \to G,\ g \to p g$. I write $G$ additively, so $pg$ means $g+g+\ldots+g$ ($p$ times).

  2. Introduce subgroups $G_i = \ker \varphi^i$. Clearly, $G_0 = 1$. Also, $G_1 \simeq \mathbb{Z}_p$ is the unique subgroup of order $p$. Since $G$ is finite, $G = G_i$ for large enough $i$. Let $n$ be the minimal such $i$.

  3. Note that $\varphi$ induces an injective homomorphism from $G_{i+1}/G_i$ to $G_i/G_{i-1}$ for every $i \geq 1$. It follows using induction that $G_{i+1}/G_{i} \simeq \mathbb{Z}_p$ for every $0 \leq i < n$.

  4. Now it is quite easy to see that any element from $G_n \setminus G_{n-1}$ generates $G_n$, so $G=G_n$ is cyclic.

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