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.

`$H\times\{0\}^{k-1}$`

, namely the isomorphism sending`$(g_1,g_2,g_3,\dots,g_k)$`

to`$(g_1,g_2-g_1,g_3-g_1,\dots,g_k-g_1)$`

. – Andreas Blass Oct 22 '12 at 11:32