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This is probably a stupid question, but, per this post Group presentation of $A_5$ with two generators, a presentation for $A_5$ is given by $A_5 \cong \langle x,y \mid x^5=y^2=(xy)^3=1 \rangle$. In this post Group presentation for semidirect products, it is shown how to make a presentation by generators and relators for semi-direct product if one knows a presentation for the quotient group, the kernel group, and the outer action of the quotient group on the kernel group.

It is known that the binary icosahedral group $B$ is a group extension of $A_5$ by $\mathbb{Z}_2$, $1 \to \mathbb{Z}_2 \to B \to A_5 \to 1$. How could one make a presentation by generators and relators for $B$ using the presentations $\mathbb{Z}_2 \cong \langle z \mid z^2=1 \rangle$ and $A_5 \cong \langle x,y \mid x^5=y^2=(xy)^3=1 \rangle$?

In general, if $1 \to K \to E \to Q \to 1$, how could one make a presentation by generators and relators for $E$ using the presentations for $K$ and $Q$ if one knew the outer action of $Q$ on $K$ and one had a set-theoretic section $\sigma: Q \to E$ which gave rise to a factor set $[,]: Q \times Q \to Z(K)$ which induced the correct element of $H^2[Q; Z(K)]$ for the group extension, per the following paper?

https://math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Ho.pdf

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If the extension is non-split, then relations $r=1$ of $Q$ become $r=w_K(r)$ for some word $w_K(r)$ in the generators of $K$, which have to be determined. You could in principal calculate $w_K(r)$ from the cocycle of the extension.

In your example, we can take $C_2 = \langle z \mid z^2 = 1 \rangle$. In fact $w_K(r)=z$ for each relation $r=1$ of $A_5$ works (but there are other possible choices). Then you get $$\langle x,y,z \mid x^5=y^2=(xy)^3=z, z^2=1, xz=zx, yz=zy \rangle$$ as a presentation of the binary icosahedral group. In fact the relations $xz=zx$ and $yz=zy$ are redundant, because $z$ is a power of both $x$ and of $y$, so you can leave those out if you want to.

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  • $\begingroup$ Super-helpful!! Thank you so much, @DerekHolt! (The last two relators are, of course, the slide relators, which normally wouldn't be able to be omitted in a group extension with a less trivial kernel group.) $\endgroup$ Commented Aug 19, 2021 at 23:32
  • $\begingroup$ Just one more basic question: you wouldn't happen to know a published reference for this (book or paper; website?) for this? $\endgroup$ Commented Aug 20, 2021 at 17:37
  • $\begingroup$ Sorry to have disturbed. I think this webpage mathoverflow.net/questions/44631/… and Lemma 2.1 of this paper arxiv.org/abs/1003.5117 answer the question. $\endgroup$ Commented Aug 20, 2021 at 17:47
  • $\begingroup$ Ah yes, and I see that I answered the mathoverflow question myself! $\endgroup$
    – Derek Holt
    Commented Aug 20, 2021 at 18:40
  • $\begingroup$ I didn't even notice that! LOL $\endgroup$ Commented Aug 21, 2021 at 2:37

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