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I'm reading a text on group theory and I've encountered a theorem I can't understand. Here are the statement and the proof (translated from Polish). I believe there must be some mistakes in it, only I don't know how many. I'm changing nothing in the symbols used.

Let $H_i$ be subgroups of $G$. Then $$G=\bigcup_{i=1}^{n}\bigcup_{j=1}^{n_i}H_ia_{ij}\implies\exists_i[G:H_i]<\infty.$$

And the proof is this:

By induction over $n.$ For $n=1$ it is clear because


and so $[G:H_1]\leq n_1.$ Now the step. Let $[G:H_n]=\infty.$ There exists a coset $H_ng$ which is contained in


and so

$$H_na_{n_j}\subset \bigcup_{i=1}^{n-1}\bigcup_{j=1}^{n_i}H_ia_{ij}g^{-1}a_{n_j}.$$

We treat the element $a_{ij}g^{-1}a_{n_j}$ as $b_{i,j,t},$ where $t$ corresponds to the pair $(n,j).$ Therefore

$$\bigcup_{j=1}^{n_n}H_n a_{n_j}\subset\bigcup_{i=1}^{n-1}\bigcup_{j=1}^{n_t}H_ib_{ijt}. $$



and the inductive hypothesis gives us that one of these cosets has a finite index.

Here ends the proof. I understand very little of this. Most importantly, in the statement of the theorem, $n,$ $n_i$ and $a_{ij}$ are nowhere defined, so I don't even know what is being proven. Secondly, in the proof, there is a number of further undefined symbols. What could $s$ be? What is $a_{n_j}?$ What does it mean that $t$ corresponds to the pair $(n,j)?$ What does it mean that a coset has a finite index? I know that a subgroup may have a finite index, but a coset?

Could you please explain to me what the theorem says?

share|cite|improve this question
For the setup: There are $n$ subgroups $H_1, \ldots, H_n$, and for each $i = 1, \ldots, n$ we're taking $n_i$ right cosets of $H_i$ having representatives $a_{i1}, \ldots, a_{in_i}$. The $a_{n_j}$ does seem to come out of nowhere, though. Are you sure about the transcription? – Dylan Moreland Feb 10 '12 at 15:34
@DylanMoreland Yes, I'm sure. – user23211 Feb 10 '12 at 15:40
The theorem says that if $G$ is finite union of cosets of finitely many subgroups $H_i$, then for some $j$, $H_j$ has finite index. This is a particular case of a theorem of Neumann, whose proof I believe you'll enjoy more:… – user641 Feb 10 '12 at 15:49
It's Lemma 4.1, sorry. – user641 Feb 10 '12 at 15:50
up vote 7 down vote accepted

Here's a rephrasing of the theorem (which I believe is due to Bernhard Neumann):

If we can write a group $G$ as a finite union of cosets of subgroups, then at least one of the subgroups has finite index in $G$.

In other words, if $K_1,\ldots,K_m$ are subgroups of $G$, not necessarily distinct, and $g_1,\ldots,g_m$ are elements of $G$ such that $$G = K_1g_1\cup\cdots\cup K_mg_m,$$ then at least one of the $K_i$ is of finite index in $G$.

Now, to prove this, let us assume that $G$ is a union of cosets, as written. Since the same subgroup may appear more than once, we will rewrite the union by putting equal subgroups together. So let us say that there are actually only $n$ distinct subgroups that appear, and call them $H_1,\ldots,H_n$; in other words, $H_i\neq H_j$ if $i\neq j$, and for each $i$, $1\leq i\leq m$, there exists $j$ such that $K_i=H_j$. We rename the $g_j$ as well, by indexing them with two indices, one relating to the subgroup, one to the coset in the subgroup. That is, $$\begin{align*} G &= K_1g_1\cup\cdots\cup K_mg_m\\ &= H_1a_{11} \cup H_1a_{12}\cup\cdots\cup H_1a_{1n_1}\\ &\quad \cup H_2a_{21} \cup H_2a_{22}\cup\cdots \cup H_2a_{2n_2}\\ &\vdots\\ &\quad \cup H_n a_{n1} \cup H_na_{n2}\cup\cdots\cup H_n a_{nn_n}. \end{align*}$$ And we can write that as $$G = \bigcup_{i=1}^n \bigcup_{j=1}^{n_i} H_ia_{ij}.$$

(I haven't done anything, I've just rewritten the "$G$ is a finite union of cosets of subgroups" into the same form as the statement you have).

The proof is going to be done by induction on the number of distinct subgroups that show up (this is what $n$ counts).

If $n=1$, then we are writing $G$ as a finite union of cosets of a single subgroup. Since the cosets of $H_1$ partition $G$, and $G$ equals the union of $n_1$ cosets, then the index of $H_1$ in $G$ is at most $n_1$ (there could be repeats among the cosets); but in any case, $[G:H_1]\leq n_1$.

Now assume that the result holds when there are strictly less than $n$ distinct subgroups involved, and that $G$ is a union of finitely many cosets of $n$ subgroups. Look at $H_n$. If $[G:H_n]$ is finite, we are done: at least one of the subgroups has finite index. So let us assume that $[G:H_n]=\infty$.

Because $H_n$ has infinite index, no finite union of cosets of $H_n$ can equal $G$, or even "miss" only finitely many elements of $G$; so $$H_na_{n1}\cup\cdots\cup H_na_{nn_n}$$ must "miss" infinitely many cosets of $H_n$ in $G$. In particular, there exists $g\in G$ such that $H_ng\cap (H_na_{n1}\cup\cdots\cup H_na_{nn_n})=\varnothing$. Since $$G = \bigcup_{i=1}^n \bigcup_{j=1}^{n_i} H_ia_{ij},$$ it must be the case that $$H_ng \subseteq \bigcup_{i=1}^{n-1}\bigcup_{j=1}^{n_i} H_ia_{ij}.$$ Multiplying on the right by $g^{-1}a_{nj}$ we get $$H_na_{nk} \subseteq \bigcup_{i=1}^{n-1}\bigcup_{j=1}^{n_i} H_i(a_{ij}g^{-1}a_{nk})$$ for $k=1,\ldots,n_n$.

For simplicity, rename $a_{ij}g^{-1}a_{nk}$ as $t_{ijk}$. Then $$H_na_{n1}\cup\cdots \cup H_na_{n_n} \subseteq \left(\bigcup_{i=1}^{n-1}\bigcup_{j=1}^{n_i}H_it_{ij1}\right) \cup\cdots\cup \left(\bigcup_{i=1}^{n-1}\bigcup_{j=1}^{n_i} H_it_{ij(n-1)}\right).$$ That means that we can write $G$ as: $$\begin{align*} G &= \bigcup_{i=1}^n\bigcup_{j=1}^{n_i} H_ia_{ij}\\ &= \left(\bigcup_{i=1}^{n-1}\bigcup_{j=1}^{n_i} H_ia_{ij}\right)\cup \left(\bigcup_{j=1}^{n_n}H_na_{ij}\right)\\ &= \left(\bigcup_{i=1}^{n-1}\bigcup_{j=1}^{n_i} H_ia_{ij}\right)\\ &\qquad \cup \left(\bigcup_{k=1}^{n_n}\bigcup_{i=1}^{n-1}\bigcup_{j=1}^{n_i} H_it_{ijk}\right). \end{align*}$$ But this expresses $G$ as a finite union of cosets of $n-1$ distinct subgroups. By the induction hypothesis, at least one of the subgroups $H_1,\ldots,H_{n-1}$ must be of finite index in $G$, which is what we wanted to prove.

Note: I find some of the notation unfortunately: I would have used $m_i$ as the number of cosets of $H_i$, to prevent the ugly "$n_n$"...

share|cite|improve this answer
Thank you, Arturo, this is a fantastic answer. – user23211 Feb 10 '12 at 20:39
@ymar: I think Neumann probably gave a different proof (can't remember the details, and I haven't looked at the link Dylan provided), which would likely be better; but I tried to stick to the one you presented. – Arturo Magidin Feb 10 '12 at 20:40
I have looked at the link and the proof is essentially the same as far as I can tell. – user23211 Feb 10 '12 at 20:47

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