I'm reading about elliptic curve groups and the structure of the finite subgroups. Suppose I have a finite abelian group $T$ of order $n=p^k m$ where $p$ is a prime that does not divide $m$, and also $T\subseteq \mathbb{Z}_n\oplus \mathbb{Z}_n$. Then we decompose $T=G\oplus P$ where $P$ is the (unique) Sylow $p$-subgroup, and $\vert G\vert =m$. We also know that $G\subseteq \mathbb{Z}_m\oplus \mathbb{Z}_m$ (maybe this follows directly from the structure of $T$; it can also be concluded from elliptic curve facts). Then the proof I'm reading concludes that $T$ can be decomposed as $T_1\oplus T_2$ where $\vert T_1\vert$ divides $\vert T_2\vert$, and $p$ divides only $\vert T_2\vert$.
I can see how we can decompose $G=G_1\oplus G_2$ and so $T=G_1\oplus G_2\oplus P$. If $P$ is a cyclic group (though I don't see why it should be), then $G_2\oplus P$ can be rewritten as a cyclic subgroup since their orders are co-prime. But I don't know why we would conclude that this new cyclic group has order divisible by the order of $G_1$, or how to decompose it in some other way.
Maybe I could decompose each subgroup into Sylow subgroups as follows: $$G_i = P_{i1}\oplus\dots P_{i\ell_i}\oplus Q_{i1}\oplus\dots\oplus Q_{i \ell_0}$$ for $i\in \{1,2\}$, where $P_{ij}$ and $Q_{ij}$ are the Sylow subgroups for some primes $p_j$ that divide $\vert G_i\vert$, such that $\vert P_{ij}\vert = p_j^{a_j}$ for a prime $p_j$ that divides only one of $\vert G_1\vert$ or $\vert G_2\vert$, and $\vert Q_{ij}\vert=q_j^{a_j}$ for a prime $q_j$ that divides both $\vert G_1\vert$ and $\vert G_2\vert$.
Then since the order of $P_{ij}$ (for every $i$ and $j$) is co-prime to over other group in the decomposition, I can rewrite: $$\begin{align} G_1\oplus G_2 = &P_{11}\oplus\dots P_{1\ell_1}\oplus Q_{11}\oplus\dots\oplus Q_{1 \ell_0} \oplus P_{21}\oplus\dots\oplus P_{2\ell_2}\oplus Q_{21}\oplus\dots\oplus Q_{2\ell_0}\\ = &\left(Q_{11}\oplus\dots\oplus Q_{1 \ell_0}\right) \oplus \left(P_{11}\oplus\dots P_{1\ell_1}\oplus P_{21}\oplus\dots\oplus P_{2\ell_2}\oplus Q_{21}\oplus\dots\oplus Q_{2\ell_0}\right)\\ = & G_1' \oplus G_2'\end{align}$$
such that $G_1'$ and $G_2'$ are cyclic. Now, all primes that divide $\vert G_1'\vert$ will also divide $\vert G_2'\vert$; however, I have no guarantee about the power of each prime that divides the two orders.
(if it helps, the proof is Corollary 7.4 in these notes.)
I assume that you meant for both $T_i$ to be cyclic, else I can take $T_1=1$ and $T_2=T$. If this is the case, the proof is nothing to do with elliptic curves, and simply a statement about finite abelian groups.
Any abelian group $G$ of order $m$ can be decomposed as a direct sum of cyclic groups $$G=G_1\oplus G_2\oplus \cdots \oplus G_r.$$ As you have already noted, you can already decompose everything in sight into Sylow $q$-subgroups for all the $q$ dividing $m$, so I will just treat one such $q$. A direct sum of cyclic $q$-groups can be ordered so that each factor has larger order than the last. But for you, since everything is a subgroup of $m\oplus m$, there are at most two terms. Thus each Sylow $q$-sugbroup of $G$ can be decomposed as $G_{q,1}\oplus G_{q,2}$, where $G_{q,i}$ is cyclic and $|G_{q,1}|\leq |G_{q,2}|$. But these are $q$-groups, so the latter inequality implies that the one divides the order of the other.
Now take $T_1$ to be the sum of the $G_{q,1}$, and $T_2$ to tbe the sum of $G_{q,2}$ and $P$. It remains to check that each $T_i$ is cyclic. But the direct sum of cyclic groups of coprime order is cyclic, so we are done.