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?

enter image description here

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.