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How can I find a non-abelian subgroup in this Cayley table? I tried all kinds of things, like $\{A,B,C,D,E,F,G,H\}$ or $\{A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W\}$, but they are all abelian. Maybe the trick is to get some blocks that are coloured together, but then I have to extend these blocks, and I just seem to get too many elements then.

Any ideas?

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  • $\begingroup$ Do you need a proper subgroup? Otherwise you can just take the whole thing. $\endgroup$ – Eric Wofsey May 30 '17 at 23:59
  • $\begingroup$ @Eric oh yea, I forgot to mention that! $\endgroup$ – Sha Vuklia May 31 '17 at 0:02
  • $\begingroup$ What's wrong with $\{A,B,\ldots,X\}$? It's closed under the operation, so it's a subgroup, and it's not abelian. For example $XV\neq VX$. $\endgroup$ – verret May 31 '17 at 9:17
  • $\begingroup$ $\{A,B,\ldots,H\}$ will also do, as for example, $HF\neq FH$. $\endgroup$ – verret May 31 '17 at 9:19
  • $\begingroup$ By the way, assuming this is indeed the Cayley table for a group, then $\{A,\ldots,H\}$ is the quaternion group. (Non abelian of order $8$, with a unique involution.) And then $\{A,\ldots,X\}$ is $SL(2,3)$ I believe. ($I$ has order $3$, and doesn't commute with any element of order $4$ in the quaternion group, which I think can only happen in $SL(2,3)$. Actually, an easier argument: one can see that the Sylow 2-subgroup is normal.) My guess would then be that the full group is $GL(2,3)$ (but I didn't check). $\endgroup$ – verret May 31 '17 at 9:35
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Notice that $oi=D$ and $io=F$. (I may have reversed these, not sure which order to read the table in. The important thing is they don't commute. Edit: fixed reversal) Can you figure out what the subgroup generated by $o$ and $i$ is? I.e. what is the subgroup $\langle o,i\rangle\le G$? Alternatively, find any subgroup containing both of them.

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  • $\begingroup$ That’s going to be a huge subgroup, no? Because we have (it’s in the reverse order!) $oi=D,oD=r,or=K,oK=a,oa=T,oT=j,\cdots$, and this just keelson going. Is this only way? Because it’s going to take me ages, even if I don’t make mistakes along the way. $\endgroup$ – Sha Vuklia May 31 '17 at 0:09
  • $\begingroup$ Unfortunately, you can't generate a non-abelian subgroup with just one element. ($\langle \sigma\rangle$ is a cyclic group, so abelian). So you have to contain two non-commuting elements, and the smallest subgroup containing them is the subgroup generated by them. I just wrote down the first ones I saw in the table; maybe if you look more carefully you will find two that generate a smaller subgroup. Alternatively, you could try to find some subgroup containing both of them (this might be easier to see with the coloring as you mentioned). $\endgroup$ – helloworld112358 May 31 '17 at 0:38
  • $\begingroup$ Also, as mentioned above, you can take the whole group if a proper subgroup isn't required. $\endgroup$ – helloworld112358 May 31 '17 at 0:39
  • $\begingroup$ It has to be a proper subgroup. Right now I'm just trying out all kinds of combinations, without a strategy really.. I hope I find a non-abelian subgroup in the end:l $\endgroup$ – Sha Vuklia May 31 '17 at 0:44
  • $\begingroup$ I think I see a way to compute this somewhat efficiently, will edit my answer $\endgroup$ – helloworld112358 May 31 '17 at 0:47

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