This question appears in my textbook's exercises, who can help me prove it?
If a group $G$ is the set-theoretic union of finitely-many cosets, $$G=x_1S_1\cup\cdots\cup x_nS_n$$ prove that at least one of the subgroups $S_i$ has finite index in $G$.
I think that the intersection of these cosets is either empty or a coset of the intersection of all the $S_i$. I want to start from this point to prove it. So I suppose none of these $S_i$ has finite index, but I don't know how to continue?
For completeness, the Neumann proof is roughly as follows.
Let $r$ be the number of distinct subgroups in $S_1,S_2,\dots,S_n$. We will prove by induction on $r$.
If $r=1$, then $G=\bigcup_{i=1}^n x_iS_1$ and $S_1$ thus has finite index in $G$.
Now assume true for $r-1$ distinct groups, and assume $S_1,\dots,S_n$ has $r>1$ distinct groups, with $S_{m+1}=\cdots=S_n$ the same, and $S_i\neq S_n$ for $i\leq m$.
If $G=\bigcup_{i=m+1}^n x_iS_i$ then $S_n$ has finite index, from the $r=1$ case.
So assume $h\not\in \bigcup_{i=m+1}^n x_iS_i$. Then $hS_n\cap \left(\bigcup_{i=m+1}^n x_iS_i\right) = \emptyset$, since $hS_n$ is a distinct coset of $S_n$.
So $$hS_n \subseteq \bigcup_{i=1}^m x_iS_i$$ So for any $x\in G$, $$xS_n=xh^{-1}hS_n\subseteq \bigcup_{i=1}^m xh^{-1}x_iS_i$$
So we can replace $x_iS_i=x_iS_n$ for $i>m$ with a finite number of cosets of $S_1,\dots,S_m$. And now we have $G$ as a finite union of cosets of the $r-1$ distinct subgroups in $S_1,\dots,S_m$. And, by induction, one of those must have finite index in $G$.
(The harder result, that some $S_i$ has finite index smaller than $n$, does not follow from this argument, since we might increase the number of terms when we do the reduction from $r$ to $r-1$.)
I believe that this is a nontrivial result. It was proven in
B. H. Neumann, Groups covered by finitely many cosets, Publ. Math. Debrecen 3 (1954), 227–242. MR 17, 234.
Unfortunately, I could not find the original argument. Perhaps someone with a sharper Google-fu can retrieve it.
(I am curious to know what book you found this in. It seems like a cruel exercise, at least assuming that your book is an introductory book on group theory.)
This question in general (especially given a bit of research) seems very nontrivial. So I assume that what the book intends is that $G$ is the union of finitely many cosets which also happen to be subgroups. Otherwise, it would be difficult to define what one means by finite index as $|G:S|$ usually implies that $S$ is a subgroup of $G$ at the very least. So I will help you show this for $S_i$ being subgroups, my notation is that $S_i=g_iH_1$ for some subgroup $H_i$ and some $g_i \in G$.
It is natural to prove this by induction with some thought. Let $n$ be the number of subgroups you have broken $G$ into. We also assume that the $S_i$ are distinct (if they are not, this is easy to take care of). First case is $n=1$. Then $G=S_1$. So it is clear that $S_1$ has finite index (being 1).
Now assume that the statement is true for $n\geq 1$. Let consider $\{S_i\}_{i=1}^{n+1}$, the union of whose elements is all of $G$. No here is where I leave steps to you.
First, suppose that $\{S_1,\cdots,S_n\}$ is the same as a different cover $\{H_1,\cdots,H_{n+1}\}$, where each of the $H_i$ are distinct. If $|G:H_{n+1}|$ is finite, does the result follow? Why?
Now suppose that $|G:H_{n+1}|=\infty$. First, rewrite the union, $\cup_{i=1}^{n+1}S_i$ as $$G=\cup_{i=1}^kS_i\bigcup \cup_{j=1}^m H_i$$where $k=n+1-m$ with each of the $S_i \in \{H_1,\cdots,H_m\}$ for at least one of the subscripts $i=1,\cdots,k$. (This was the tough part) Now what does $|G:H_{n+1}|=\infty$ imply? Can you find an $aH_{n+1} \neq x_j H_{n+1}$ for all possible $x_j$?. Then what does that say about $aH_{n+1} \cap x_jH_{n+1}$? Then use the fact that $aH_{n+1} \subset G$ to write it as a finite union of subsets (specifically the $S_i$). Moreover, $x_jH_{n+1}$ are then all finite unions of the $S_i$. Then $G$ is a finite union of all the $H_i$. The result then follows from applying the induction hypothesis.
This is a not very lengthy, but thought intensive process. Just take it slowly and it is very doable.
