A problem with lemma about cyclic groups I have a small, but important for me, question about one lemma. I cannot understand one thing.
Lemma: Let us suppose that $G$ is a finite, abelian group which is not cyclic and for each prime $p_{i}$, $p_{i} \mid ord(G)$, $x_{i} \in G$ and $l_{i} \in \mathbb{N}$, such that $ord(x_{i}) = p_{i}^{l_{i}}$ and $l_{i}$ is maximal for $p_{i}$. THEN: There exist $N$ such that $N = p_{1}^{l_{1}} \cdot p_{2}^{l_{2}} \cdot ... \cdot p_{k}^{l_{k}} \mid ord(G), N < ord(G)$ and for each $x \in G ~~ x^{N} = 1$.
I understand, why $N < ord(G)$ but I cannot prove why for each $x \in G ~~ x^{N} = 1$. My teacher said me, that we can assume that there exist $g \in G$ such that $N_{1} = ord(g) \nmid N$. Then there exist $l \in \mathbb{N}$, that $p^{l} \mid N_{1}$ but $p^{l} \nmid N$. He claimed that it means that it is equivalent to the fact that there exist $j \in \mathbb{N}$, that $p=p_{j}$ and $l > l_{j}$ what is in conflict with the fact that $l_{j}$ is maximal for $p_{j}$. I don't understand it. Could you explain me this fact? 
Thank you very much for your help.

A finite abelian group is a direct product of its Sylow subgroups, and if they were all cyclic, $G$ would also be cyclic, so one of them, say the $p$-Sylow subgroup, is non-trivial and non-cyclic.  Then if $Ord(G)=p^nq_1^{a_1}\cdot\cdot\cdot q_r^{a_r}$, set $N=p^{n-1}q_1^{a_1}\cdot\cdot\cdot q_r^{a_r}$.  This works, since the elements in the first coordinate will have order a divisor of $p^n$ but less than $p^n$ since the p-Sylow is not cyclic.
To see that a finite abelian group is cyclic if all its Sylow subgroups are cyclic, let there be $s$ factors in the representation of $G$ as a direct product of its cyclic subgroups, take the $s$-tuple containing a cyclic generator of each factor, and show that it generates $G$.
Also, note that the result is only true for abelian groups.  For example, the non-abelian group of order 21 has all Sylow subgroups cyclic, but 21 is the smallest exponent $N$ such that $x^N=1$ for all $x$ in the group.
