Ok, this has been bugging me for a while, and I'm sure there's something obvious I'm missing. The references I've looked at for this result in an effort to resolve the issue didn't address it.

$G$ is a group, $\mathbb{Z}[G]$ its integral group ring, $I_G$ the augmentation ideal (i.e. the kernel of the map $\mathbb{Z}[G]\rightarrow\mathbb{Z}$ sending all group elements to 1), and $G/G'$ is the abelianization of $G$. I'm attempting to show that $I_G/I_G^2$ is isomorphic to $G/G'$.

It's straightforward to show that $$(g-1)+(h-1)\equiv(gh-1)\bmod I_G^2\hskip0.5in(*)$$ and thus every equivalence class mod $I_G^2$ is equal to some $(g-1)+I_G^2$.

Now all we have to do is define a map $\phi:G/G'\rightarrow I_G/I_G^2$ and show that it has an inverse $\psi:I_G/I_G^2\rightarrow G/G'$. The definitions are obvious enough: $$\phi(gG')=(g-1)+I_G^2$$ $$\psi((g-1)+I_G^2)=gG'$$ and starred equation shows that these are homomorphisms. But! My problem is showing that these maps are well-defined. For example, if $(g-1)+I_G^2=(h-1)+I_G^2$, i.e. $$(g-1)-(h-1)\equiv (gh^{-1}-1)\equiv0\bmod I_G^2,$$ we need to show that $$\psi((g-1)+I_G^2)=gG'=hG'=\psi((h-1)+I_G^2),$$ i.e. $gh^{-1}\in G'$.

If we have $gh^{-1}\in G'$, i.e. $gh^{-1}$ equals some $\prod_{a,b\in G} (aba^{-1}b^{-1})^{n_{a,b}}$, then $$gh^{-1}-1=\left(\prod_{a,b\in G} (aba^{-1}b^{-1})^{n_{a,b}}\right)-1,$$ and, unwinding using the starred equation, $$\left(\prod_{a,b\in G} (aba^{-1}b^{-1})^{n_{a,b}}\right)-1\equiv \sum n_{a,b}(aba^{-1}b^{-1}-1)\equiv$$ $$\sum n_{a,b}\left[(ab-1)-(a-1)-(b-1)\right]=\sum n_{a,b}(a-1)(b-1)\equiv0\bmod I_G^2$$ but for some reason I can't make this work the other way. I'm sure I'm being silly; someone please point out where.

  • $\begingroup$ Please include the statement of the problem you are trying to do in the body of the message, not just the subject line... $\endgroup$ – Arturo Magidin Apr 13 '11 at 3:46
  • $\begingroup$ @Arturo: good point, edited. $\endgroup$ – Zev Chonoles Apr 13 '11 at 3:49
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    $\begingroup$ I'm trying to unwind what you are doing, but one possibility is that instead of trying to show a map on the quotient is well defined, you could try defining a map from $I_G$ onto $G/G'$ which does what you want, and then show the kernel is exactly $I_G^2$. That is: let the isomorphism theorems to the work for you... $\endgroup$ – Arturo Magidin Apr 13 '11 at 3:52
  • $\begingroup$ ... or, as Steve shows, not even having to show the kernel is exactly $I_G^2$, but only that it contains it; that will do it since you have the map going the other way too. $\endgroup$ – Arturo Magidin Apr 13 '11 at 4:02

You have maps $\phi:\ G\rightarrow I_G/I_G^2$ and $\psi:\ I_G\rightarrow G/G'$. $\phi$ is a homomorphism because $\phi(gh) = (gh-1) = (g-1) + (h-1) + (g-1)(h-1)$, and $(g-1)(h-1)$ lies in $I_G^2$. Since $I_G/I_G^2$ is an abelian group, $\phi$ induces a well-defined map from $G/G'$ to $I_G/I_G^2$.

Now take an element of the form $(g-1)(h-1)$ in $I_G^2$. $(g-1)(h-1) = gh-g-h+1 = (gh-1)-(g-1)-(h-1)$. Thus $\psi((g-1)(h-1)) = \psi(gh-1)-\psi(g-1)-\psi(h-1)=ghg^{-1}h^{-1}G' = [g^{-1},h^{-1}]G'=G'$, so $\psi(I_G^2) = 1$, and so $\psi$ induces a well-defined map from $I_G/I_G^2$ to $G/G'$.

  • $\begingroup$ Thanks Steve! I'd considered this approach many times, and I could have sworn that I was running into the same problems I had with my approach above, but of course now looking at it I have no idea what I was doing wrong, this is very clear and simple. $\endgroup$ – Zev Chonoles Apr 13 '11 at 5:08

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