Writing a group as a union of subgroups Let $G$ be a group and $H$ be a subgroup of $G$ of infinite index. Suppose that I can write $G$ as the union
$$G=H\cup H_1\cup\dotsb\cup H_k$$ where $H_1,\dotsc,H_k$ are subgroups of $G$. Is it the case that $H\subset \bigcup_{j=1}^{k}H_j$?
Edit: I wanted to ask for infinite index, not finite. Sorry.
I tried doing this by supposing the existence of an element $h\in H\setminus \bigcup_i H_i$. Then, there exists a sequence of element $x_n$ of the group $G$ such that the $x_n H$ are two by two distinct. It implies in particular that only one of the $x_n H$ can be equal to $H$. I do not know how to conclude.
 A: The first step of you reasoning is the good one, so I follow your idea here. I use your notations and I suppose that $h\in H\setminus \bigcup_{i =1}^{k} H_i$.
WLOG, suppose that $x_0$ is the unique element of the sequence $(x_n)_{n\geq 0}$ such that $x_nH=H$.
For any $n\geq 1$, there exists $i$ such that $x_n\in H_i$. Since there are finitely many $H_k$, there exists $i_1$ such that $$x_n\in H_{i_1}\text{ for infinitely many } n.$$ We can therefore extract a subsequence $(x_n^{(1)})_{n\geq 0}$ such that $x_n^{(1)}\in H_{i_1}\setminus H$ and such that if $m\neq n$, then $$x_n^{(1)}H\neq x_m^{(1)} H.$$ Our goal now is to create a similar sequence, but of elements in $H_{i_1} \cap H_{i_2}$ for some $i_2\neq i_1$, and continue like this until we have all the $H_i$'s in the intersection, which will lead to a contradiction.
The sequence defined by $y_n = hx_n^{(1)}$ is neither in $H$, nor in $H_{i_1}$, since $h\notin \cup_i H_i$ and $x_n^{(1)}\notin H$.
Using the same reasoning as before, we can find $i_2 \neq i_1$ and extract from $(y_n)$ a sequence $(y_{\varphi (n)})_{n\geq 0}$ such that, $$y_{\varphi(n)}\in H_{i_2}\text{ for every n.}$$
We now need to get into $H_{i_1}$ without leaving $H_{i_2}$ using this new sequence. The only way to do this is to translate by a well chosen element of $H_{i_2}$.
To this end, define a new sequence $x_n^{(2)}$ by $$x_n^{(2)} = (y_{\varphi(0)})^{-1}y_{\varphi(n)}.$$ By definition of $y_{\varphi(n)}$, we have that $x_n^{(2)}\in H_{i_2}$. Moreover, $y_{\varphi(0)} = hx_p^{(1)}$ and $y_{\varphi(n)} = h x_q^{(1)}$ for some integers $p,q$, by definition. Consequently, $$x_n^{(2)} = (x_p^{(1)})^{-1}x_q^{(1)}\in H_{i_1},$$ and $x_n^{(2)}H \neq x_m^{(2)} H$ if $m\neq n$. Indeed, otherwise we would have $x_r^{(1)}H = x_q^{(1)} H$ where $r$ is such that $y_{\varphi(m)}=hx_r^{(1)}$.
We have therefore constructed a sequence $(x_n^{(2)})_{n\geq 0}$ of elements of $H_{i_1} \cap H_{i_2}$ such that the $x_n^{(2)}H$'s are two by to distincts. This is what I announced. We can continue this procedure until we are left with a sequence $x_n^{k}$ in $H_{i_1}\cap...\cap H_{i_k}=\cap_{i =1}^{k}H_i $ such that the $x_n^{(k)}H$'s are two by two distincts. Let $N$ be such that $x_N^{(k)}\in G\setminus H$. Then, $hx_N^{(k)}$ cannot be in $H$, and neither can it be in $\cup_{i }H_i$. This implies that $G\neq H\cup H_1\cup...\cup H_k$, which is the desired contradiction.
