Given an abelian group `$G$` and subgroup `$H$`, `$G^k/\Delta \cong G/H \times G^{k-1}$`, where `$\Delta = \{ (a, ..., a)\mid a\in H \}$`. Why is this the case?

2012-10-22T07:07:46Z

It looks like the natural way to define this isomorphism is $(g_1, ..., g_k )\mapsto ([g_1], g_2, ..., g_k ) $ where $[g_1]$ is the congruence class of $g_1$ in $G/H$ . I can see why this is onto, but I don't think $\Delta$ would be part of the kernel. So, it looks like I'm considering the wrong map, but what else is there to consider?

Thanks! I ran across this in a research paper, and it's just not coming to me why this is true.

Given an abelian group `$G$` and subgroup `$H$`, `$G^k/\Delta \cong G/H \times G^{k-1}$`, where `$\Delta = \{ (a, ..., a)\mid a\in H \}$`. Why is this the case? - MathOverflow

Given an abelian group `$G$` and subgroup `$H$`, `$G^k/\Delta \cong G/H \times G^{k-1}$`, where `$\Delta = \{ (a, ..., a)\mid a\in H \}$`. Why is this the case?