# Exponent of a direct product of cyclic groups

I have an answer to a homework question that I am not sure is correct. The question is show that if $G \cong C_{n_1} \times C_{n_2} \times \cdots \times C_{n_k}$ for positive integers $n_1, n_2, \cdots, n_k$, then $e(G) = \text{lcm}(n_1, n_2, \cdots, n_k)$, where $e(G)$ is the exponent of the group $G$ (defined in the question as the greatest order of any element in the group).

I know how to show that $o((a_1, a_2, \cdots, a_k)) = \text{lcm}(n_1, n_2, \cdots, n_k)$ where the $a_i$ are generators of the $C_{n_i}$, but then can I just claim that because each $a_i$ has the greatest order of any element in the group $C_{n_i}$ (since $a_i$ is a generator), then the element $(a_1, a_2, \cdots, a_k)$ has the greatest order in $C_{n_1} \times C_{n_2} \times \cdots \times C_{n_k}$? I just wanted to make sure of this last part.

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If I understand the question, yes, if you can show that $o((a_1, a_2, \cdots, a_k)) = \text{lcm}(n_1, n_2, \cdots, n_k)$ - provided the $a_i$ represent the generators of the respective factors - you can then use your justification:

...because each $a_i$ has the greatest order of any element in the group $C_{n_i}$ (since $a_i$ is a generator), then the element $(a_1, a_2, \cdots, a_k)$ has the greatest order in $C_{n_1} \times C_{n_2} \times \cdots \times C_{n_k}$.

More formally: Let $L=\text{lcm}\,(n_1,n_2,...,n_k)$.

Show that $\forall x\in G$, $\; [x^L=e]$. And so $e(G)\leq L$.

Thus $e(G)\geq o(a_1,a_2,...,a_k)=L$. And so we have $e(G)=L$

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+1 I hope the OP cam show that what you noted italically. – Babak S. Jan 19 '13 at 17:52
@amWhy Thanks, my point was whether I can just state that or show more formally, I like that more formal argument. – user50229 Jan 19 '13 at 17:58

Let $L=lcm(n_1,n_2,...,n_k)$. Verify that $\forall x\in G [x^L=e]$. Thus it follows that $e(G)\leq L$. Finally, $|(a_1,a_2,...,a_k)|=L$. Hence $e(G)\geq |(a_1,a_2,...,a_k)|=L$. Hence, $e(G)=L$

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