Index of Directed Union of Groups

96 Views Asked by At

Let $I$ be a nonempty directed set, and $H_i \le G_i$ both be directed sequence of groups living in some ambient group. Is there a nice formula for $|\bigcup_{i \in I} G_i : \bigcup_{i \in I} H_i|$, perhaps involving a limit or supremum of the $|G_i : H_i|$? I must confess, I don't really have much intuition for what the index should be.

1

There are 1 best solutions below

0
On

I suspect that the best we can say is that $[\cup G_i:\cup H_i]\leq \mathrm{sup}\{[G_i:H_i]\mid i\in\mathbb{N}\}$.

Indeed, let $g_1,\ldots,g_n$ represent distinct cosets of $\cup H_i$ in $\cup G_i$. Then there exists an index $N$ such that $g_1,\ldots,g_n\in G_N$, and since $g_jH_N\subseteq g_j\cup H_ik$, it follows that the the $g_i$ represent distinct cosets of $H_N$ in $G_N$. Thus, $n\leq [G_N:H_N]\leq \sup\{[G_i:H_i]\mid i\in\mathbb{N}\}$.

But it is easy to come up with all kinds of examples where the inequality is strict. For example, we take subgroups of the Prüfer $p$-group whose union is the whole thing, and subgroups of them whose union is also the whole thing, but the indices are either constant, or even growing. So the supremum is whatever we want, from $1$ through $\infty$, but $[\cup G_i:\cup H_i]=1$. Simple tweaks give you $[\cup G_i:\cup H_i] = p^i$ for any $i$, even if the indices grow without bound, or are always $p^{i+1}$, etc.