All elements of $N := \oplus_{p \in \mathcal{P}} \mathbb{Z}/p\mathbb{Z}$ are of finite order

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I'm trying to show the statement in the title (where $\mathcal{P}$ is the set of prime numbers, and $N$ is viewed as a $\mathbb{Z}$-module). The idea is that every element of $N$ has only finitely many nonzero entries, so their order is at most their least common multiple. However, take the element $n = (0 \mod 2, 2 \mod 3, 0\mod 5, 4 \mod 7, 0 \mod p)_{p \in \mathcal{P}}$. Note that lcm(2,4) = 8.

Then

$8n = 8(0 \mod 2, 16 \mod 3, 0\mod 5, 32 \mod 7, 0 \mod p)_{p \in \mathcal{P}} = (0 \mod 2, 1 \mod 3, 0\mod 5, 4 \mod 7, 0 \mod p)_{p \in \mathcal{P}} \neq 0_N$.

I had expected to get $0_N$, so where is my reasoning faulty?

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The order of a finite family of elements with finite order is the l.c.m. of the orders of each, not the l.c.m. of the elements.