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This concerns the updated question. The answer is yes.

If $G$ is represented as the union of a finite number of subsets $A_1, \dots,A_n$, then one of the subsets generates a finite index subgroup of $G$. Indeed, let $G_i$ be the subgroup which is generated by $A_i$. Then, $G$ is the union of the $G_i$.

B.H. Neumann proved that if $G$ is the union of a finite number of left cosets of subgroups, then one of the subgroups is of finite index (see this postthis post). In particular, there exists some $i$ such that $G_i$ is of finite index.

This concerns the updated question. The answer is yes.

If $G$ is represented as the union of a finite number of subsets $A_1, \dots,A_n$, then one of the subsets generates a finite index subgroup of $G$. Indeed, let $G_i$ be the subgroup which is generated by $A_i$. Then, $G$ is the union of the $G_i$.

B.H. Neumann proved that if $G$ is the union of a finite number of left cosets of subgroups, then one of the subgroups is of finite index (see this post). In particular, there exists some $i$ such that $G_i$ is of finite index.

This concerns the updated question. The answer is yes.

If $G$ is represented as the union of a finite number of subsets $A_1, \dots,A_n$, then one of the subsets generates a finite index subgroup of $G$. Indeed, let $G_i$ be the subgroup which is generated by $A_i$. Then, $G$ is the union of the $G_i$.

B.H. Neumann proved that if $G$ is the union of a finite number of left cosets of subgroups, then one of the subgroups is of finite index (see this post). In particular, there exists some $i$ such that $G_i$ is of finite index.

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Andreas Thom
Andreas Thom
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This concerns the updated question. The answer is yes.

If $G$ is represented as the union of a finite number of subsets $A_1, \dots,A_n$, then one of the subsets generates a finite index subgroup of $G$. Indeed, let $G_i$ be the subgroup which is generated by $A_i$. Then, $G$ is the union of the $G_i$.

B.H. Neumann proved that if $G$ is the union of a finite number of left cosets of subgroups, then one of the subgroups is of finite index (see this post). In particular, there exists some $i$ such that $G_i$ is of finite index.