Let $G=\oplus_{k=0}^n\mathbb{Z}a_k$ and $H=<a_0,2a_1-a_0,2a_2-a_1,...,2a_n-a_{n-1}>\subset G$

I need to prove that $G/H\cong \mathbb{Z}/2^n\mathbb{Z}$, but I can't do it. Does anyone have any idea?

  • 1
    $\begingroup$ Try to use $G/H \simeq (G/K)/(H/K)$ (assuming everything makes sense). It will be better if you try for $n=1,2,3$ to get the idea before going for the general case. $\endgroup$
    – Krish
    Oct 22, 2017 at 17:01


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