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I was playing a bit with SAGE to study transitive and regular groups, to head towards Mathieu groups. Actually, I've generated a permutation group like this:

$$ G = \lt(1,2,3)(5,6)\gt $$

This turns out to be a 6 order permutation group, i.e:

$$ G = \{e, (5,6), (1,2,3), (1,3,2), (5,6)(1,2,3), (5,6)(1,3,2)\} $$

This group is manifestly intransitive (and so non regular), since showing 3 orbits:

$$ O = \{\{1, 2, 3\}, 4, \{5,6\}\} $$

So far, so good. But it came to my mind to wonder about which group G is isomorphic to. Actually, from the classification of groups which order is a product of two primes, I know the 6 order groups fall into 2 distinct isomorphism classes:

1) Cyclic Group $C6$

2) Symmetric group $S3$ (or equivalently dihedral group $D6$)

However, inspecting visually the G elements, we can see immediately there's only one inversion, i.e. (5, 6), while $S3$ has two of them, of course.

Just to have further confirmation on that, I've tried to query SAGE for isomorphisms and it correctly pointed out there were not the case.

So, and I know this may sound a bit a dumb question:

What is this 6 order G Group in term of isomorphism? It's for sure a subgroup of $S6$, but I'm more interested to see where my reasoning is wrong in trying to find an isomorphism which I cannot find evidently.

Thank for your support, as always.

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Your group contains an element of order $6$, namely $(5 \; 6)(1\; 2 \; 3)$. That already shows that is is a cyclic group of order $6$, i.e $G \cong C_6$.To find an isomorphism to some other cyclic group $H$ of order $6$, just send the generator of $G$ to a generator of $H$.

If you were trying to define an isomorphism to $S_3$ you would always fail because isomorphisms preserve orders. Thus, as you realized, the problem would be that $G$ only contains one element of order $2$.

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  • $\begingroup$ you are right indeed @ThorWittich. Actually I retried in SAGE setting T = CyclicPermutationGroup(6) and then it told me they are isomorphic, correctly. Actually you are right: since that element has order 6, it generates a subgroup of order 6 and, since that subgroup matches the order of the group, it must match it isomorphically. Don't knowwhat I was doing wrong before, thanks a lot for your help. $\endgroup$ – riccardoventrella Oct 26 '19 at 8:58
  • $\begingroup$ Glad that I could help. $\endgroup$ – TMO Oct 26 '19 at 9:24

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