# Abelianization of a semidirect product

I believe there is a straightforward formula for the abelianization of a semi-direct product: if $G$ acts on $H$, and we form the semi-direct product of $G$ and $H$ in the usual way, and the abelianization of this semi-direct product is the product $G^{ab}\times (H^{ab})_{G}$.

(Here the subscript $G$ denotes taking the coinvariants with respect to $G$. That is, $(H^{ab})_{G}$ is a the quotient of $H^{ab}$ by the subgroup generated by elements of the form $h^g-h$ for $h$ in $H$ and $g$ in $G$, and where the superscript $g$ denotes the action of $G$ on $H^{ab}$ induced by the action of $G$ on $H$.)

Does anyone happen to know a good reference for this?

• Does it really need a reference? Write down the presentation for the semi-direct product, that gives you a presentation matrix for the abelianization and it's pretty much immediate from there, no? – Ryan Budney Aug 16 '10 at 3:16
• That's what I thought. However, a referee requested that I explain the formula; it seems that giving a reference is more appropriate than explaining the thing in detail. (I'm nervous about only explaining it very briefly, given that referee made an especial request for clarification...) – blt Aug 16 '10 at 3:21
• If you don't find a reference, just write a one-paragraph explanation along the lines of Ryan's comment. If it is a mathematics journal, it should be sufficient. – Victor Protsak Aug 16 '10 at 3:41
• It is a mathematics journal, for a research paper in number theory (not a textbook). Given the weight of the consensus here, I will write a short explanation along the lines of Greg's below. Thank-you all for giving me the confidence to do so! – blt Aug 16 '10 at 12:37

I agree with Ryan and Victor, except that you don't need presentations. The subgroup $[G \ltimes H,G \ltimes H]$ is generated by $[H,H] \cup [G,H] \cup [G,G]$, so you can write $$(G \ltimes H)^{ab} = (G \ltimes H) / \langle [H,H] \cup [G,H] \cup [G,G] \rangle.$$ If you apply the relators $[H,H]$, you get $G \ltimes H^{ab}$; then if you apply the relators $[G,H]$, you get $G \times (H^{ab})_G$; then finally if you apply $[G,G]$, you get $G^{ab} \times (H^{ab})_G$. You can add this as an extra half-paragraph or footnote rather than giving a citation.
• You don't even need to mention commutators: any homomorphism from $G\ltimes H$ to an abelian group factors through $G\times H^{ab}$; then through $G\times (H^{ab})_G$; finally through $G^{ab}\times (H^{ab})_G$. – mephisto Apr 21 '11 at 21:32