# 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? [closed]

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.

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## closed as off-topic by Stefan Kohl, Ricardo Andrade, Chris Godsil, abx, Felipe VolochNov 16 at 17:38

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This is not really a research level question, but set $\Delta(G)$ to be the corresponding diagonal subgroup of $G^{k}$ using $G$ in place of $H$. Embed $G^{k-1}$ inside $G^{k}$ by adding the identity in the $k$-th component. Note that $G^{k} = \Delta(G) \times G^{k-1}$ with these identifications. –  Geoff Robinson Oct 22 '12 at 8:09
A slight modification of Geoff's answer may be easier to see. There is an isomorphism from $G^k$ to itself taking $\Delta$ to $H\times\{0\}^{k-1}$, 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
Thanks. I really like Andreas' answer, but I unfortunately can't designate it as "best answer" for some reason (perhaps a problem with my account?). –  Reeve Oct 22 '12 at 18:59