Let $G$ be a finite, abelian group, prime $p\mid |G|$ and $G$ contain $p-1$ elements order of $p$. Prove $G$ is cyclic. 
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.
 A: 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.
