Let G be a group and let F be the family of all subgroups of finite index. Let {xn} be a sequence in G. We define this sequence to be Cauchy if given H∈F there exists n0 such that for m,n≥n0 we have xnxm−1∈H. We define multiplication of two sequences componentwise.Now, we define {xn} to be a null sequence if given H∈F there exists n0 such that for n≥n0 we have xn∈H.Prove that the null sequences N form a normal subgroup of the Cauchy sequences C.

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Brendon Stein · Accepted Answer

Let&amp;nbsp;C&amp;nbsp;be the group of Cauchy sequences and&amp;nbsp;N&amp;nbsp;the group of null sequences of elements&amp;nbsp;∈G.&amp;nbsp;Let&amp;nbsp;F◃&amp;nbsp;be the set of all normal subgroups of finite index in&amp;nbsp;G. Clearly&amp;nbsp;F◃⊆F. Every&amp;nbsp;H∈F&amp;nbsp;gives rise to a an action of&amp;nbsp;F&amp;nbsp;on the cosets: a homomorphism&amp;nbsp;F→∑&amp;nbsp;&amp;nbsp;(FH). The group on the right is finite, hence the kernel&amp;nbsp;N&amp;nbsp;of this action is&amp;nbsp;∈F, and clearly&amp;nbsp;N≤H. For every&amp;nbsp;H∈F, we find&amp;nbsp;N∈F◃&amp;nbsp;with&amp;nbsp;N≤H.&amp;nbsp;Thus,&amp;nbsp;{xn}&amp;nbsp;is Cauchy only if for every&amp;nbsp;N∈F◃, there exists&amp;nbsp;n0&amp;nbsp;such that&amp;nbsp;xnxm−1∈N∀n,m≥n0and{xn}&amp;nbsp;is null only if for every&amp;nbsp;N∈F◃, there exists&amp;nbsp;n0&amp;nbsp;such that&amp;nbsp;xn∈N∀n,m≥n0For&amp;nbsp;N∈F◃, we have a homomorphism&amp;nbsp;ϕN:C→GN:&amp;nbsp;Given&amp;nbsp;{xn}∈C, pick&amp;nbsp;n0&amp;nbsp;such that&amp;nbsp;xnxm−1∈N∀n,m≥n0&amp;nbsp;and set&amp;nbsp;ϕN({xn})=xn0N.&amp;nbsp;This is well-defined. By combining all&amp;nbsp;ϕN, we obtain a homomorphismϕ:C→∏N∈F◃GNWe see that&amp;nbsp;kerϕ=N. Therefore,&amp;nbsp;N◃C.

Let G be a group and let F

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