# Show that $[u,g]\in G'$ where $u\in N_U(G)$

Let $G=\langle H,g\rangle$ where H is an abelian subgroup of index $2$. Let $\Bbb{Z}G$ be the group ring and $u$ be a unit of $\Bbb{Z}G$ which normalizes $G$. Then we can write $u$ as $\alpha_1+\alpha_2g$ where $\alpha_1,\alpha_2 \in \Bbb{Z}H$.

Now my question is, how do we realize commutator $[u,g]$ lying in derived group $G'$.

I am doing a paper on Normalizer problem and in there it just says note that $[u,g]\in G'$. I dont see it that obvious.

• But the derived group is generated by all commutators?! Or have I misunderstood the question? – M.U. Dec 28 '15 at 19:30
• Yes thats true. But $u$ is sum of elements from G. How does [u,g] opens up and written as product of commutators – Bhaskar Vashishth Dec 28 '15 at 19:36

Let $u \in N_{U(\mathbb{Z}G)}(G)$, then ${conj}(u)$ is an automorphism of the group $G$. Firstly, we show that ${conj}(u)$ is class-preserving. Let $C$ be a conjugacy class of $G$. Define in $\mathbb{Z}G$ the element $$e_C = \sum_{g \in C}{g}.$$ Then, for any $h \in G$, we get that $$h e_C = \sum_{g \in C}{hg} = \sum_{g \in G}{hgh^{-1}h} = e_C h.$$ This shows that $e_C$ is a central element in $\mathbb{Z}G$. Hence, $$conj(u)(e_C) = u^{-1} e_C u = e_C.$$ Hence, $conj(u)(C) \subseteq C$.
Hence, for any $g \in G$ there exists $h \in G$ such that $u^{-1}g^{-1}u = h^{-1}g^{-1}h$. Moreover, this shows that $$[u,g] = u^{-1}g^{-1} u g = h^{-1}g^{-1} h g = [h,g] \in [G,G]$$