# 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? 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. 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.

I don't think that the referee has the right to demand a longer explanation than this, unless maybe you are writing a textbook.

• 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 $G^{ab}\times (H^{ab})_G$. Apr 21 '11 at 21:32

A description of the derived subgroup of a semidirect product, from which the abelianization can be obtained, was published in:

Daciberg Lima Gonçalves, John Guaschi The lower central and derived series of the braid groups of the sphere Trans. Amer. Math. Soc. 361 (2009), 3375-3399. http://www.ams.org/journals/tran/2009-361-07/S0002-9947-09-04766-7/ (Proposition 3.3)

You may also find it in their preprint: http://arxiv.org/abs/math/0603701 (Proposition 29)

If you have the semidirect product $$H\rtimes G$$ then you have the next group split extension

$$1\rightarrow H\rightarrow H\rtimes G\rightarrow G\rightarrow 1$$,

We have the Hochschild–Serre spectral sequence where $$\mathbb{Z}$$ is a trivial $$H\rtimes G-$$module

$$E^{2}_{p,q}=H_{p}(G,H_{q}(H,\mathbb{Z}))\Rightarrow H_{p+q}(H\rtimes G,\mathbb{Z})$$,

Since the the map $$H\rtimes G\rightarrow G$$ is a split surjection, it follows that the map (edge morphism)

$$H_{n}(H\rtimes G,\mathbb{Z})\rightarrow H_{n}(G,\mathbb{Z})=E^{2}_{n,0}$$ is a slit surjection and thus $$E^{2}_{n,0}=E^{\infty}_{n,0}$$

In particular we have that the diferenttial $$d:E^{2}_{2,0}\rightarrow E^{2}_{0,1}$$ is zero (since $$E^{2}_{2,0}=E^{\infty}_{2,0}$$). Therefore $$E^{2}_{0,1}=E^{\infty}_{0,1}$$.

It follows that there is a exact sequence

$$0\rightarrow E^{\infty}_{0,1}\rightarrow H_{1}(H\rtimes G,\mathbb{Z})\rightarrow E^{\infty}_{1,0}\rightarrow 0$$

which splits, in this case we have that

$$H_{1}(H\rtimes G,\mathbb{Z})=E^{\infty}_{0,1}\times E^{\infty}_{1,0}$$

Note that

$$H_{1}(H\rtimes G,\mathbb{Z})=(H\rtimes G)^{Ab}$$

$$E^{\infty}_{1,0}=G^{Ab}$$

and

$$E^{\infty}_{0,1}=H_{0}(G, H^{Ab})=(H^{Ab})_{G}$$

From this we have the result by using spectral sequences.