$d_sg = 0$ for all $g$ in a (nontrivial) finite abelian group $G$

Question

We know for a finite nontrivial abelian group $G$ there exist positive integers $1 < d_1|...|d_s$ such that

$G \cong \bigoplus_{i=1}^{s} \mathbb{Z}/ d_i\mathbb{Z}$ and $|G| = d_1...d_s$

Now, I need to show that in such a group

$\forall g \in G \ \ d_sg = 0$, that is, $\forall g \in G \ \ |g|$ divides $d_s$.

By Lagrange's theorem $|g|$ divides $|G| = d_1...d_s$. Note that $d_i$'s doesn't have to be prime.

Answer 1

I'll expand on the hint of user26857 in the comments (since it's better if the question is to be officially answered in case someone will search for it and I'm not sure if he wants to write an answer). I thank him for the following idea.

So, we have that $G \cong \bigoplus_{i=1}^{n} \mathbb{Z}/ d_i \mathbb{Z}$. To each $g \in G$ corresponds an (unique) element $([k_1]_{d_1}, ..., [k_s]_{d_s}) \in \bigoplus_{i=1}^{n} \mathbb{Z}/ d_i \mathbb{Z}$ send by an isomorphism.

These elements have the same order. And $$d_s([k_1]_{d_1}, ..., [k_s]_{d_s}) = (m_1d_1[k_1]_{d_1}, m_2d_2[k_2]_{d_2}, ..., d_s[k_s]_{d_s}) = ([0]_{d_1}, ..., [0]_{d_s} ),$$ so $d_s$ divides the order of $g$, hence, $d_sg = 0$.