I was reading this post, and I was wondering if you had instead the direct sums as such:
$$0\rightarrow\mathbb{Z}_{p^{a_1}}\oplus...\oplus\mathbb{Z}_{p^{a_n}}\rightarrow\mathbb{Z}_{p^{b_1}}\oplus...\oplus\mathbb{Z}_{p^{b_n}}\rightarrow\mathbb{Z}_{p^{c_1}}\oplus...\oplus\mathbb{Z}_{p^{c_n}}\rightarrow0,$$
you could still reach similar conclusions that $\sum b_i=\sum a_i+\sum c_i$ and $b_1\leq a_1+c_1$, like how the answer in the link would have instead that $j+(m+n-j)=m+n$ and $j\leq m+n$. If this is true, could it be proved similarly?
To take this out of the comment section.
The answer to your first question is “yes”. This is because if $$1 \to A\to B \to C\to 1$$ is an exact sequence of groups (using multiplicative notation for generality as this does not require groups to be abelian, hence the use of $1$ instead of $0$), this implies that $B$ contains a normal subgroup $M\triangleleft B$ such that $A\cong M$ and $B/M\cong C$. In particular, if at least two of the groups in question are finite then so is the third, and $|B|=|A||C|$. This gives the equality you have.
The answer to your second question is “not as stated, at any rate.”
This is because you are not saying anything about the maps. Even if you assume the groups are given in “ascending order”, that is, with $x_1\leq x_2\leq\cdots\leq x_n$ (where $x$ stands for either $a$, $b$, or $c$), you cannot simply assume that the maps you have will respect the decomposition.
To give an explicit example, consider $$ 0 \to \mathbb{Z}_3\oplus\mathbb{Z}_9 \stackrel{f}{\to} \mathbb{Z}_{27}\oplus\mathbb{Z}_{27} \stackrel{g}{\to}\mathbb{Z}_3\oplus\mathbb{Z}_9\to 0$$ where the maps are given as follows:
These are morphisms, as is easy to verify; $f$ is one-to-one, since $(3b,9a)=(0,0)$ implies that $9|b$ and $3|a$, hence $(a,b)=(0,0)$ in $\mathbb{Z})3\oplus\mathbb{Z}_{27}$; $g$ is clearly surjective. The composition $g\circ f$ is zero, since $g(f(a,b)) = g(3b,9a) = (3b\bmod 3,9a\bmod 9) = (0,0)$. Thus, $\mathrm{Im}(f)\subseteq \mathrm{ker}(g)$.
Finally, if $(x,y)\in\mathrm{ker}(g)$, then $3|x$ and $9|y$, so we can write $x=3x’$, $y=9y’$, and $0\leq x’\lt 9$, $0\leq y’\lt 3$; hence $(x,y)=f(y’,x’)$. Thus, $\mathrm{ker}(g)=\mathrm{Im}(f)$, proving that the sequence is exact.
However, $b_1=3$, $a_1=c_1=1$, so $b_1\not\leq a_1+c_1$. Thus, the requested inequality does not hold.
The problem is that even if you require that the summands be listed in ascending order of size, there is nothing structural that requires the maps $f$ and $g$ to respect the coordinate order.
That’s why you can’t seem to extend the argument: because it’s false. You cannot conclude the stuff you want about $a_1,b_1,c_1$ themselves just from knowing you have an exact sequence.
Now, it looks like you just decided to repost this question but adding the condition that the summands go in descending order, so the example above does not work for that condition.
Under that condition, the inequality does hold. To see this, note that if $A=\mathbb{Z}_{p^{a_1}}\oplus\cdots\oplus\mathbb{Z}_{p^{a_n}}$, $B=\mathbb{Z}_{p^{b_1}}\oplus\cdots\oplus\mathbb{Z}_{p^{b_m}}$, and $C=\mathbb{Z}_{p^{c_1}}\oplus\cdots\oplus\mathbb{Z}_{p^{c_k}}$, with $a_1\geq\cdots\geq a_n$, $b_1\geq \cdots \geq b_m$, $c_1\geq \cdots \geq c_k$, and $$ 0 \to A\to B\to C\to 0$$ is an exact sequence (note here we do not even require the same number of summands in each of the groups), then the exponent of $A$ is $p^{a_1}$, the exponent of $B$ is $p^{b_1}$, and the exponent of $C$ is $p^{c_1}$ (the exponent being the smallest positive integer $r$ such that $rx=0$ for every $x$ in the group; and in these cases, there is an element that achieves the bound). In particular, if $x\in B$, then $p^{c_1}x$ must map to the trivial element in $C$, and therefore must be in the image of $A$; in particular, $p^{a_1}(p^{c_1}x)=0$, because every element of $A$ is of exponent $p^{a_1}$. Thus, $p^{a_1+c_1}x=0$ for all $x\in B$, which means that $p^{b_1}\leq p^{a_1+c_1}$, and hence $b_1\leq a_1+c_1$.