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Let $G$ be a finite Abelian group: $G=\mathbb{Z}_{n_1}\times\cdots\times\mathbb{Z}_{n_k}$ and $n_k \mid n_{k-1} \mid \cdots \mid n_1$.

Show that $G$ contains an element of order $m$ iff $m\mid n_1$.

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    (cont'd) is just a case of a new user not knowing what is expected rather than laziness. But you have to **prove** that you are not just lazy, otherwise the helpful answers will stop. I hope I don't sound too harsh. At the moment you are (hopefully accidentally rather than knowingly) broadcasting such an impression, and it is only fair of me to warn about the negative effects this practice has. Proof? Look at the tally of downvotes your most recent questions have raked in!2012-12-05

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Exercise 1. Show that, for any finite groups $G$ and $H$,

$\mathrm{ord}_{G\times H}(a,b)=\mathrm{lcm}\big(\mathrm{ord}_G(a),\mathrm{ord}_H(b)\big).$

Proceed further to show that $\mathrm{ord}_{G_1\times\cdots\times G_k}(a_1,\cdots,a_k)=\mathrm{lcm}\big(\mathrm{ord}_{G_1}(a_1),\cdots,\mathrm{ord}_{G_k}(a_k)\big)$.

Exercise 2. Show that the $\mathrm{lcm}$ of any number of divisors of $n_1$ will divide $n_1$.

Now suppose you have a tuple in $\Bbb Z_{n_1}\times\cdots\times\Bbb Z_{n_k}$. The order (in $\Bbb Z_{n_j}$) of the $j$th component will divide $n_j$ by Lagrange's theorem, and by transitivity of the $\mid$ relation will also divide $n_1$. By exercise one, the order of the tuple itself in the product will be the lcm of these component orders, and by exercise two will divide $n_1$. Conversely, for every divisor $d|n_1$, $\Bbb Z_{n_1}$ will have an element of order $d$, which can easily be transplanted into the product group (put it inside the first coordinate and make all the other coordinates zero).