If $H$ is a subgroup of infinite index and $G = H \cup H_1 \cup H_2 \cup \cdots \cup H_p$, show that $G = H_1 \cup H_2 \cup \cdots \cup H_p$. 
Let $G$ be a group, $H$ a subgroup of infinite index (that is there exists a sequence $(x_n) \in G^\mathbb{N}$ of distinct elements such that $\forall i \neq j, x_i H \neq x_j H$), $H_1, \ldots, H_p$ subgroups of $G$ such that $G = H \cup H_1 \cup H_2 \cup \cdots \cup H_p$, show that $G = H_1 \cup H_2 \cup \cdots \cup H_p$.

By posting this exercise, I am looking for different approches to this problem.
What I've done so far (draft):

*

*The property verified by $(x_n)$ is equivalent to $\forall i\neq j, x_i^{-1} x_j \notin H$. So if $x_0 \in H$ therefore $\forall n \ge 1, x_n \notin H$. Let's say that if $\exists i, x_i \in H$ we take $i=0$.

*I studied the case $p=1$ with $G= (\mathbb{R}, +), H = \mathbb{Z}, (x_n) = (\frac{2}{3})^n$ but my proof isn't generalizable.

*For $p=1$, we have $G = H_1 \cup H$. Suppose $H_1 \neq H \cup H_1$ ie. there exists $h \in H - H_1$. We have $x_1 \notin H$ so $x_1 \in H_1$, so $h x_1 \notin H_1$, so $hx_1 \in H$, so $x_1 \in H$ which is absurd. I tried to use the same method for $p=2$ but it's harder: suppose there exists $h \in H-H_1-H_2$, we still have $x_1 \notin H$ so let's say $x_1 \in H_1$, therefore $hx_1 \notin H_1$ so $hx_1 \in H_2-H_1-H$...

*Back to the general case: Using 1. we see that $\forall n \ge 1, \exists j, x_n \in H_j$ so there exists $i_1$ such that $H_{i_1}$ contains an infinite number of images of the sequence $(x_n)$. Let's call $(x^{(i_1)}_n)$ a sequence of distinct terms such that $\forall n \ge 1,x^{(i_1)}_n \in H_{i_1}$. Suppose there exists $h \in H-H_1 - \cdots - H_p$, we see that $h x^{(i_1)}_n \in H-H_{i_1}$ so by induction we can find a sequence $(y_n)$ of distinct terms such that $\forall n, y_n \in H-H_1 - \cdots - H_p$.

Reference: Exercise 2.13 in Exercices de mathématiques: oraux X-ENS (Algèbre I), by Francinou, Gianella, and Nicolas.
 A: One of the subgroups $H_i$, say $H_1$, must contain infinitely many of the $x_i$, which we can label as $y_1,y_2,\ldots,$.
Then, choosing $h \in H \setminus (H_1 \cup \cdots \cup H_p)$, we have  $hy_i \not\in H_1 \cup H$ for all $i$, and so some other $H_i$, say $H_2$, must contain infinitely many of them, say $hz_1,hz_2,\ldots,$
Note that $z_i^{-1}z_j \in (H_1 \cap H_2) \setminus H$ for all $i \ne j$, so $hz_1^{-1}z_j \not\in H_1 \cup H_2 \cup H$ for all $j>1$, and hence some other $H_i$, say $H_3$ contains infinitely many of them, say $hz_1^{-1}w_1, hz_1^{-1}w_2.\ldots$.
Then $w_i^{-1}w_j \in (H_1 \cap H_2 \cap H_3) \setminus H$ for all $i \ne j$ so $hw_1^{-1}w_j \not\in H_1 \cup H_2 \cup H_3 \cup H$ for all $j>1$, and hence some other $H_i$, say $H_4$ contains infinitely many of them, etc. etc.
A: This result follows from:

B.H. Neumann's Lemma.
Suppose $G$ is a group and $G=C_1\cup\ldots\cup C_n$ where each $C_i$ is a coset of a subgroup $H_i\leq G$. The $G$ is the union of just the $C_i$ for which $H_i$ has finite index.

I'm going to give a proof of this result using some elementary notions from combinatorial number theory. Let $G$ be a group. A subset $A$ of $G$ is called syndetic if $G$ can be covered by finitely many left translates of $A$. A subset $A$ is called piecewise syndetic if $A\cup Z$ is syndetic for some non-syndetic $Z$. The basic fact is:
Fact 1: Given $A,B\subseteq G$, if $A\cup B$ is piecewise syndetic, then one of $A$ or $B$ is piecewise syndetic.
Proof: By assumption there is a non-syndetic set $Z$ such that $A\cup B\cup Z$ is syndetic. So if $B\cup Z$ is not syndetic then $A$ is piecewise syndetic (by definition); and if $B\cup Z$ is syndetic then $B$ is piecewise syndetic (by definition).
Note that a subgroup of a group is syndetic if and only if it has finite index. The next fact strengthens this to piecewise syndetic. Given subsets $A,B$ of a group $G$, let $AB$ denote product set $\{ab:a\in A,b\in B\}$.
Fact 2: If $C$ is a left coset of a subgroup $H\leq G$, and $C$ is piecewise syndetic, then $H$ has finite index. 
Proof: Suppose $C\cup Z$ is syndetic for some non-syndetic set $Z$. Then $G=F(C\cup Z)$ for some finite set $F$. So $G=FC\cup FZ$. If $G=FC$ then the proof is done. Otherwise $FC$ is a proper subset of $G$, and so $FZ$ contains some left coset of $H$, which can write as $gC$ for some $g\in G$.  Then $FC=(Fg^{-1})gC\subseteq (Fg^{-1})FZ$, and thus $G=(Fg^{-1}F)Z\cup FZ$, which is a contradiction since $Z$ is not syndetic.
Proof of B.H. Neumann's Lemma. Suppose $G=C_1\cup \ldots \cup C_n$, where each $C_i$ is a coset of a subgroup $H_i\leq G$. Note that any right coset $Hg$ of a subgroup $H$, is a left coset $g(g^{-1}H g)$ of the conjugate subgroup $g^{-1}Hg$, which has the same index as $H$. So, without loss of generality, we may assume each $C_i$ is a left coset.
Let $X$ be the union of all $C_i$ such that $H_i$ has finite index, and let $Y$ be the union of all $C_i$ such that $H_i$ has infinite index. So we have $G=X\cup Y$ and we want to show $G=X$. Toward a contradiction, suppose $G\neq X$. We first claim that $Y$ contains a coset of a finite index subgroup of $G$. (Throughout the proof, when I say "coset" I mean left coset.) If $X=\emptyset$ then $G=Y$ and this is obvious. Otherwise, $X$ is a union of cosets of a finite index subgroup $K$ (namely, let $K$ be the intersection of all $H_i$ of finite index). So since $G=X\cup Y$, but $G\neq X$, it follows that $Y$ contains a coset of $K$.
Now, $Y$ is syndetic since it contains a coset of $K$, and $K$ has finite index. So $Y$ is piecewise syndetic (take $Z=\emptyset$). But recall that $Y$ is the union of all $C_i$ such that $H_i$ has infinite index. So by Fact 1, one of these $C_i$'s is piecewise syndetic. But this contradicts Fact 2.
