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 most recent 30 from http://mathoverflow.net 2013-05-21T17:40:32Z http://mathoverflow.net/feeds/question/110303 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110303/given-an-abelian-group-g-and-subgroup-h-gk-delta-cong-g-h-times-g 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? Reeve 2012-10-22T07:07:46Z 2012-10-22T07:07:46Z <p>It looks like the natural way to define this isomorphism is <code>$(g_1, ..., g_k )\mapsto ([g_1], g_2, ..., g_k )$</code> where <code>$[g_1]$</code> is the congruence class of <code>$g_1$</code> in <code>$G/H$</code>. I can see why this is onto, but I don't think <code>$\Delta$</code> would be part of the kernel. So, it looks like I'm considering the wrong map, but what else is there to consider?</p> <p>Thanks! I ran across this in a research paper, and it's just not coming to me why this is true.</p>