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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?

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    $\begingroup$ 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? $\endgroup$ Aug 16 '10 at 3:16
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    $\begingroup$ 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...) $\endgroup$
    – blt
    Aug 16 '10 at 3:21
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    $\begingroup$ 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. $\endgroup$ Aug 16 '10 at 3:41
  • $\begingroup$ 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! $\endgroup$
    – blt
    Aug 16 '10 at 12:37
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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.

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

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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.

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