4

enter image description here

I was asked to construct a Cayley table for this Cayley diagram and it occurred to me that it will be impossible.

My reasoning is that at first glance this appears to be the group $D_4$. It has to be, since it has only two generators. But now, the group $D_4$ has 5 self inverses. This will be impossible to achieve with the Cayley diagram above since we have $$ea=I$$ and $$cg=I$$ In the group $D_4$ there is only one unequal pair that are inverses of each other, here there are two. So you can't generate 5 self inverses. Perhaps this is easier to see if you construct the identity skeleton for $D_4$ and this graph.

So I was wondering where have I gone wrong? Is this graph in fact erroneous in that it doesn't represent an actual group?

Thanks.

DLV
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    $D_4$ is not the only 8-element group with two generators. The diagram indicates two generators of order 4, and the two generators don't commute, so that narrows it down to just one possibility. – Gerry Myerson Oct 10 '15 at 23:25
  • So you identify a dihedral group if its generators commute? – DLV Oct 10 '15 at 23:28
  • I don't know enough theory, so are you referring to some specific "other" possibility which is a well-known result? Thanks. – DLV Oct 10 '15 at 23:29
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    $D_4 is not commutative, so its generators don't commute. But it has only two elements of order 4, and they don't generate it. So, yes, there is one other well-known group of order 8. But you don't need to know that in advance. The diagram gives you all the information you need to construct the group table. – Gerry Myerson Oct 10 '15 at 23:34
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    fwiw, I just noticed the Wikipedia article https://en.wikipedia.org/wiki/Cayley_graph contains the graph for $D_4$. – James S. Cook Oct 11 '15 at 00:43
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    Also, see http://www.weddslist.com/groups/cayley-plat/index.html for more cool examples of these. – James S. Cook Oct 11 '15 at 00:53
  • @JamesS.Cook Nice, I'll check it out. – DLV Oct 11 '15 at 00:54
  • I have done the Cayley table and its not documented on any webpage online. The group is nonabelian and its most similar is C4XC2 (which is abelian). Anyone knows whats going on? Thanks. – DLV Oct 11 '15 at 04:32
  • It is nonabelian and has three self inverses and the generators $a,g$ are of order 4. – DLV Oct 11 '15 at 05:13
  • There is no nonabelian group of order 8 with 3 self-inverse elements. – Gerry Myerson Oct 11 '15 at 05:23
  • So, the elements are $I,a,b=ag,c=a^2g,d=a^2,e=a^3,f=a^3g,g$. Now you ought to be able to construct the group table from that. E.g., $cf=caaag=a$. Why not do it, and post it as an answer? – Gerry Myerson Oct 11 '15 at 05:29
  • Should I post my Cayley table even if its that nonabelian group with 3 self-inverse elements? I've already checked my table a lot of times now. – DLV Oct 11 '15 at 05:33
  • I should probably say 4 self-inverse elements since I wasn't counting the identity as a self-inverse. (when I say 4 I am counting the identity) – DLV Oct 11 '15 at 05:35
  • http://i.imgur.com/VKJr18F.jpg That's the link to the Cayley table I've done. – DLV Oct 11 '15 at 05:39
  • i've solved my problem now, thanks for all your help. – DLV Oct 11 '15 at 15:17
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    From your Cayley table, $(eb)eb=(ce)b=bb=I$, but also $eb(eb)=e(bc)=ee=d$, so the operation is not associative, and you don't have a group. If you have been able to produce a group table, please post it as an answer to your question. – Gerry Myerson Oct 12 '15 at 02:45

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The Cayley diagram is indeed not contradictory. It represents the Quaternion group (order 8) with the following table:

enter image description here

For more information you can see a neater table here:

http://escarbille.free.fr/group/?g=8_4c&y=8&z=0|4|1|5|2|6|3|7

DLV
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