If $G$ is a finite abelian group, then TFAE:
- $G$ is cyclic
- if $p$ divides order of $G$ then $G$ contains exactly $p-1$ elements of order $p$?
The forward direction is easy but for the other part I could not solve it without using structure theorem for finite abelian group.
Any hint will be appreciated.
Best Answer
We will prove the challenging direction, that 2. as in the OP implies 1.
Lemma 1 Let $G$ be any abelian group. For each integer $p$ that divides $|G|$, let $\nu(p)$ be the largest integer that satisfies the following: There is a cyclic subgroup $H(p)=\langle \alpha \rangle$ of $G$ such that $|H(p)|$ is divisible by $p^{\nu(p)}$. Then let $p_1,\ldots, p_r$ be the distinct primes that divide $|G|$. Then there is a cyclic subgroup $H'$ such that $p_1^{\nu(p_1)}\cdots p_r^{\nu(p_r)}$ divides $|H'|$.
Indeed, to see this, for each such $p_i$, let $H'(p_i)$ be the cyclic subgroup of $H(p)$ where $|H'(p_i)|$ is precisely $p_i^{\nu(p_i)}$. Then let $\alpha_i$ be the generator of $H'(p_i)$. Then $\alpha_1\cdots \alpha_r$ has order $\prod_{i=1}^r p_i^{\nu(p_i)}$. Indeed, otherwise an equation of the form $\alpha_1^{c_1} =\prod_{i=2}^r p_i^{c_i}$ would hold for some $c_1$ such that $p_1^{\nu(p_1)}$ does not divide $c_1$; raising both sides of this to the $\prod_{i=2}^r p_i^{\nu(p_i)}$-th power would give the equation $p_1^{C_1}=e$ for some integer $C_1$ such that $p_1^{\nu(p_1)}$ does not divide $C_1$, which is impossible. $\surd$
So now, let us assume that $G$ satisfies 2. as in the OP. First, using Lemma 1, let $p_1,\ldots, p_r$ be distinct primes that divide $|G|$, and let $H'$ be a cyclic subgroup of $G$ such that $p_1^{\nu(p_1)}\cdots p_r^{\nu(p_r)}$ divides $|H'|$. Let $\alpha'$ be an element in $G$ such that $\langle \alpha' \rangle = |H'|$. For ease of notation, let $M'$ be the integer satisfying $M'=|H'|$. If $H'=G$ then we are done. So let us assume that $H'$ is a proper subgroup of $G$, and thus there is a $\beta \in G \setminus H'$. Now, as $H'$ is cyclic and it is easy to see that 1. of the OP implies 2. of the OP, there are exactly $p_i-1$ elements of $H'$ of order $p_i$, for each $i=1,\ldots, r$. So we finish showing that 2. implies 1. in the OP by constructing from $\beta$ an element $\gamma \in G \setminus H'$ such that $\gamma^{p_1} = e$, for some prime $p_1$ that divides $|G|$.
Let $N$ be the smallest positive integer such that $\beta^N=e$; so $\langle \beta \rangle$ is a subgroup with precisely $N$ elements. Then let $k$ be the smallest positive integer such that $\beta^k \in H'$, and assume WLOG that $p_1$ divides $k$, and thus $N$ as well. Then $\beta^j$ is in $H'$ iff $k|j$, and in particular only if $p_1|j$. Thus letting $\ell$ be the largest power of $p_1$ that divides $N$, it follows that $\beta^{N/p_1^{\ell}}$ is not in $H'$, because $p_1$ does not divide $N/p_1^{\ell}$ [lest there were a larger power of $p_1$ that divides $N$]. So $\ell_1$ be the largest integer that both
$p_1^{\ell_1}$ divides $N$, and
$\beta^{N/p_1^{\ell_1}}$ is not in $H'$;
there is indeed such an integer $\ell_1$. Then it follows that $\ell_1$ must be at least $2$ lest we are done. [Indeed, $\beta^N= \beta^{N/p_1^0}=e \in H'$, and if $\beta^{N/p_1}$ is not in $H'$, then set $\gamma= \beta^{N/p_1}$. Then indeed $\gamma$ is not in $H'$ and satisfies $\gamma^{p_1}=e$.] So $\beta^{N/p_1^{\ell_1-1}}$ must be in $H'$, write $\beta^{N/p_1^{\ell_1-1}}=\alpha^c$; then as $\left(\beta^{N/p_1^{\ell_1-1}}\right)^{p_1^{\ell_1-1}} = \left(\alpha^c\right)^{p_1^{\ell_1-1}}=e$, it follows that $M'$ must divide $cp_1^{\ell_1-1}$; as $p_1^{\ell_1}$ divides $N$, by Lemma 1 it follows that $p_1^{\ell_1}$ must divide $M'=|H'|$ as well, and thus, as $M'$ divides $cp_1^{\ell_1-1}$, that $p_1$ divides $c$ as well.
Thus, consider the element $\gamma=\alpha^{(M'-c)/p_1}\beta^{N/p^{\ell_1}}$. Note that $\gamma$ is not in $H'$ because $\beta^{N/p^{\ell_1}}$ is not in $H'$, and yet, $$\gamma^{p_1} = \alpha^{M'-c}\beta^{N/p_1^{\ell_1-1}}$$ $$ = \alpha^{M'-c}\alpha^c = \alpha^{M'}=e.$$ So here we constructed our desired element $\gamma$, as thus, there cannot be such an element $\beta$, and so it follows that $G$ must indeed be cyclic, and satisfy 1. of the OP after all.