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Isomorphism between $I_G/I_G^2$ and $G/G'$

Let $G$ be a finite group. Let $I\unlhd\mathbb{Z}[G]$ be the augmentation ideal of the integral group ring $\mathbb{Z}[G]$.

I'm trying to prove that

$G/G' \cong I/I^2$ as abelian groups

I defined a map $\varphi:G\rightarrow I/I^2$ by $\varphi(g)=(g-1)+I^2$. I showed that $\varphi$ is a group epimorphism. What's left is to show that $\ker\varphi=G'$. Since $\varphi$ is a homomorphism into an abelian group, we get $G'\subset \ker\varphi$.

So my question is

How to prove that $\ker\varphi \subset G'$?

My attempt at this part was to take generators for $I^2$ of the form $(g_0-1)(h_0-1)$ and show that if $g-1$ is a sum of such generators, then $g$ is in $G'$. In the case that $g-1$ is just one such generator, I could prove that actually $g=1$. I then tried to proceed by induction, but failed to do so.


marked as duplicate by Matthew Towers, Emily, Martin Argerami, Micah, tomasz Nov 21 '12 at 16:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


You have already built an epimorphism $\bar{\varphi}\colon G/G' \to I/I^2$ and you want to show that it is injective. Instead of proving that $\ker \varphi = G'$, you could build a homomorphism $\psi\colon I/I^2 \to G/G'$ such that $\psi \circ \bar{\varphi} = 1$. The existence of this $\psi$ will automatically prove that $\bar{\varphi}$ is an injection.

And it is quite easy to build such a homomorphism. Let us define $\theta:I \to G/G'$ like this: $$ \theta\left(\sum_{g \in G} n_g g\right) = \prod_{g \in G} g^{n_g} \cdot G'. $$ Note that the definition is correct, i.e. it doesn't depend on the order in which the product in the RHS is calculated, because $G/G'$ is abelian. It is clear that $\theta$ is a group homomorphism.

It is also easy to check that $I^2 \subset \ker \theta$. We can look at generators of $I^2$ that you suggested and see that $$ \theta\left((g_0-1)(h_0-1)\right) = \theta(g_0 h_0 -g_0 -h_0 +1) = g_0 h_0 g_0^{-1} h_0^{-1} \cdot G' = 1 \cdot G'. $$ And since $I^2 \subset \ker \theta$ there is an induced homomorphism $\bar{\theta} \colon I/I^2 \to G/G'$ defined by the property $\bar{\theta}(x + I^2) = \theta(x)$.

Finally, we can check that $\bar{\theta}\circ\bar{\varphi} = 1$. For any $g \in G$ we have: $$ \bar{\theta}(\bar{\varphi}(g\cdot G')) = \bar{\theta}(\varphi(g)) = \bar{\theta}((g-1)+I^2) = \theta(g-1) = g \cdot G'. $$ And since $\bar{\theta}\circ\bar{\varphi}=1$, $\bar{\varphi}$ is an injection.

  • $\begingroup$ I don't understand your first line: as the OP wrote, he only has $\,G'\subset \ker\phi\,$. How do you deduce the derived subgroup is the whole kernel? $\endgroup$ – DonAntonio Nov 21 '12 at 13:52
  • $\begingroup$ The OP has $G' \subset \ker \varphi$. This means that $\varphi$ factors through $G/G'$. That is how we get $\bar{\varphi}$. This homomorphism $\bar{\varphi}$ is an injection if and only if $G'=\ker \varphi$. Instead of proving that $G'=\ker \varphi$, I prove that $\bar{\varphi}$ is an injection. And to do that, I find a left inverse for it. $\endgroup$ – Dan Shved Nov 21 '12 at 14:01
  • $\begingroup$ @DonAntonio See my answer above (I forgot to incldue your name there). $\endgroup$ – Dan Shved Nov 21 '12 at 14:12
  • $\begingroup$ Thanks for the comment @Dan, now I understand what you meant, and +1 for the answer. $\endgroup$ – DonAntonio Nov 21 '12 at 14:36

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