# Isomorphisms Between some terms of a Long Exact Sequence

Suppose we have two long exact sequences of finite dimensional $k$-vector spaces:

$$0 \to A_1 \to A_2 \to A_3 \to \cdots$$ $$0 \to B_1 \to B_2 \to B_3 \to \cdots$$

And assume that $A_6 \cong B_6$, $A_9\cong B_9$, $A_{12} \cong B_{12}$ and so on. This is $A_{3n} \cong B_{3n}$ for $n>1$

What can I say about the other vector spaces?

You can't say much of anything in general. For instance, suppose $A_6$, $B_6$, $A_9$, $B_9$, and so on are all $0$. Then by exactness you know that (for instance) $A_7\cong A_8$ and $B_7\cong B_8$, but other than that they can be anything at all, and in particular there is no relationship between $A_7$ and $B_7$ or $A_8$ and $B_8$.
More generally, given a sequence $a_1,a_2,a_3,\dots$ of nonnegative integers, there is a long exact sequence $0\to A_1\to A_2\to A_3\to\dots$ of vector spaces with $\dim A_n=a_n$ iff for all $N$, $(-1)^N\sum_{n=1}^N(-1)^na_n\geq 0$ (this is not hard to prove by induction, using the fact that in a finite exact sequence bounded by $0$s on both sides, the alternating sum of the dimensions of the spaces is $0$). You can see that knowing the numbers $a_{3n}$ for $n>1$ puts some constraints on what the other numbers can be, but not very strong constraints, and in particular no individual other number is completely determined.