I'm trying to answer the first part of a group theory question as revision for an exam that goes as follows;

Let $G = S_3 \wr S_3$, the permutational wreath product of two symmetric groups of degree three. Give a presentation for $G$ and determine the isomorphism type of $G/[G, G]$.

I'm not sure how to go about finding generators for the wreath product itself.

Is there a method for combining the generators of the symmetric groups to form generators for the wreath prouct?

Any pointers would be much appreciated, thanks in advance!

  • 1
    $\begingroup$ If you can't find generators, then you haven't got much hope of answering the question! The obvious generating set consists of generators of the two natural subgroups isomorphic to $S_3$, giving four generators in all. For these,you could take for example: $(1,2,3)$, $(2,3)$, $(1,4,7)(2,5,8)(3,6,9)$, and $(4,7)(5,8)(6,9)$. $\endgroup$ – Derek Holt Apr 20 '17 at 18:29
  • $\begingroup$ I would like to know what a presentation of the Wreath product of two groups given by $\langle X\mid R\rangle$ and $\langle Y\mid S\rangle$ looks like in general, if possible; I might ask a question on MSE about it. $\endgroup$ – Shaun Nov 29 '18 at 23:23
  • $\begingroup$ @DerekHolt, do you know of such a presentation? I'm aware of, say, this, but, for some reason, I can't get access to it; my institution is not subscribed :( $\endgroup$ – Shaun Nov 29 '18 at 23:26
  • $\begingroup$ @shaun I have answered the question. $\endgroup$ – Derek Holt Nov 30 '18 at 8:12
  • $\begingroup$ Thank you, @DerekHolt. $\endgroup$ – Shaun Dec 1 '18 at 3:37

Here is a presentation of $S_3 \wr S_3$. Note that $a,b$ generate the top $S_3$ factor, and $x,y$ generate one of the three factors of the base group, the other two being $\langle x^a,y^a \rangle$ and $\langle x^{a^2},y^{a^2}\rangle$, where $x^a = a^{-1}xa$. $$\langle a,b,x,y,\mid a^3,b^2,(ab)^2,x^3,y^2,(xy)^2, [x,x^a],[y,x^a],[x,x^{a^2}],[y,x^{a^2}], [x,y^a],[y,y^a],[x,y^{a^2}],[y,y^{a^2}], [x,b], [y,b] \rangle.$$ A routine calculation from the presentation shows that $G/[G,G] \cong C_2 \times C_2$, but it is not hard to show that directly.

PS: In fact the four relators $[x,x^{a^2}],[y,x^{a^2}],[x,y^{a^2}],[y,y^{a^2}]$ are redundant. The fact that $\langle x,y \rangle$ commutes with $\langle x^{a^2},y^{a^2} \rangle$ follows from the fact that $\langle x,y \rangle$ commutes with $\langle x^{a},y^{a} \rangle$ by conjugating by $b$.

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