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I have the following group (at least, I think it's a group) generated by $\langle a,b,c \rangle$ where the operation $\cdot$ obeys the following rules:

  1. $a^2=b^2=c^2=1$ (where $1$ is the identity).

  2. $\cdot$ is associative.

  3. $(cb)(bc) = (bc)(cb) = 1$,

  4. $(bc)^3 = (cb)^3 = 1$,

  5. $(bc)^2 = cb$,

  6. $(bcb)^2 = 1$,

  7. $cbc = bcb$,

  8. $\forall x, \space xa = ax$.

Some of these might be redundant, which is fine, but it's obviously an issue if there's a contradiction, but I can't find one.

From these rules, I believe this is a group of order 12 with the elements $\{1,a,b,c,ab,ac,bc,cb,abc,acb,bcb,abcb\}$.

However, I've been looking at the groups of order 12, and this one doesn't seem to be isomorphic to any of them. It's not Abelian, which narrows it down to the Alternating Group, the Dihedral Group, and the Dicyclic Group.

It's not the Alternating Group, as that only has 3 elements that square to $1$, whereas my group has 8 that do ($\{1,a,b,c,ab,ac,bcb,abcb\}$).

The Dihedral Group has the right amount of elements that square to $1$, but has an order 6 element, which my group doesn't have.

I hadn't heard of the Dicyclic Group until today, and I've been having trouble finding information on it, but it also seems like it has an order 6 element.

So what am I doing wrong here?

So did I miss something here and it is actually isomorphic to one of these groups?

Is my group ill-defined to begin with?

Is it well-defined but not a group? (I'm almost sure this isn't it, because I forced the operation to be associative, and I have an identity, and everything seems to have an inverse.)

Did I miscalculate the number of elements?

Or some other mistake entirely?

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    $\begingroup$ Clearly you have $C_2\times S_3$ --- $a$ is the generator of $C_2$, $b,c$ are two different transpositions in $S_3$. By the way, you have an element of order 6, namely $abc$. $\endgroup$ – user10354138 May 13 at 17:29
  • $\begingroup$ I changed the title, because it sounded not nice. $\endgroup$ – Dietrich Burde May 13 at 18:44
  • $\begingroup$ FWIW, your rules 3 and 6 clearly follow from 1 and 2: $(cb)(bc) = cbbc = c1c = cc = 1$, $(bc)(cb) = bccb = b1b = bb = 1$ and $(bcb)^2 = bcbbcb = bc1cb = bccb = 1$. $\endgroup$ – Ilmari Karonen May 13 at 22:32
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    $\begingroup$ None of those rules actually eliminate the trivial group with one element. Of course, we're more interested in what is the most general group that satisfies all those conditions. $\endgroup$ – aschepler May 13 at 22:44
  • $\begingroup$ @aschepler I feel like when I see groups defined in this way, it's usually assumed that 2 elements are different unless some rule says so. This order 12 group is the maximal group that satisfies all these conditions, but I think there are several smaller groups, not just the trivial group, that would also satisfy them. Though I could be wrong. $\endgroup$ – RothX May 14 at 23:28
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Yes, this is a group of order 12. We already have closure, associativity, identity, we only need verification of inverses. The reason everybody has an inverse is due to the fact that $a$, $b$, and $c$ have inverses and are the generators of our group! Hence any arbitrary element, such as $abcb$ has inverse $bcba$, since multiplying by this element causes the individual $a, b,$ and $c$ terms to cancel. Now as for order 12, $\langle b,c | b^2,c^2\rangle\cong\mathbb{Z}_2\ast\mathbb{Z}_2\cong\mathbb{Z}\rtimes\mathbb{Z}_2$. We identify $\langle bc\rangle$ with $\mathbb{Z}$ and $\langle b\rangle$ with $\mathbb{Z}_2$. In addition, $\langle b,c|b^2,c^2,(bc)^3\rangle\cong\mathbb{Z}_3\rtimes\mathbb{Z}_2\cong D_3$. Hence $\langle a,b,c|b^2,c^2,(ab)^3,[a,b],[a,c]\rangle\cong\mathbb{Z}_2\oplus D_3$, which has order 12.

It was mentioned that it would be worth it to explain why $\mathbb{Z}_2\oplus D_3\cong D_6$. Note that $\mathbb{Z}_2$ represents the $0^\circ$ rotation and the $180^\circ$ rotation, and being a rotation, commutes with everything in the subgroup $D_3$ of $D_6$.

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    $\begingroup$ Good answer, but it's worth mentioning that that direct product is actually isomorphic to the dihedral group of degree 6. It's not immediately obvious, and it would answer some of the concerns in my question more specifically. $\endgroup$ – RothX May 13 at 19:27

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