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Permutations of string $ABC$ have an associated symmetry group of size $3!$. What about $AAB$ and in general? If not, is there some way to quotient out the redundant permutations from the group?

Permutations of the string $ABC$ have the group of symmetries described by $S_3$. What about $AAB$, what group?

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  • $\begingroup$ The set of all maps $f: \{1,2,3\} \to {A,B,C}$ together with the composition $(f \circ g)(x)=f(g(x))$ does not form a group: every map that is not bijection does not have an inverse element. $\endgroup$ – Zoran Loncarevic Feb 2 '17 at 17:14
  • $\begingroup$ Why are you asking about group structures if you don't know the axioms of a group? Every group element must have an inverse. There is no inverse for these functions because they aren't even bijections. $\endgroup$ – Alfred Yerger Feb 2 '17 at 17:16
  • $\begingroup$ @FruitfulApproach Well, what is a permutation, for you? And, what is permutation group $S_3$? Can you define them? $\endgroup$ – Zoran Loncarevic Feb 2 '17 at 17:18
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There is always the action of a permutation group on a string with repetitions. In tha $AAB$ case the group acts on the set $\{AAB, ABA, BAA\}$. Though we cannot associate this set with a group structure it still can be viewed as the quotient of the group by a stabilizer of one of its elements, in this example this would be the subgroup of $S3$ generated by $(1,2)$.

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    $\begingroup$ So, are you saying to take the quotient which might not be a group since the only normal subgroups are $A_n$ for $n \geq 5$? $\endgroup$ – Shine On You Crazy Diamond Feb 2 '17 at 18:00
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    $\begingroup$ Or, more generally, when the stabilizer subgroup is not normal. As another example for the string $AAAB,$ there are $4$ elements in the set. Under the action of $S_4$ the stabilizer is $S_3$, not a normal subgroup of $S_4$ . $\endgroup$ – Marc Bogaerts Feb 2 '17 at 18:30
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If we consider swapping the second and third items, you can see that sometimes that will produce a different arrangement and sometimes it will leave the arrangement unchanged - specifically, $BAA$ would not change. This would give two identity operations for this case and similarly for the other $2$ possibles arrangements.

For this particular set of strings, though, we could restrict permutations to the $3$-cycle and the group would then be valid, the cyclic group of order 3.

For a general case though, permutations of a multiset would be unlikely to have position-based permutations that could form a valid group.

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  • $\begingroup$ What do you mean by restricting to the 3-cycle ? Are you saying that $AAB \simeq ABA \simeq BAA$ has an associated group and it is the 3-cycle? $\endgroup$ – Shine On You Crazy Diamond Feb 2 '17 at 17:39
  • $\begingroup$ @FruitfulApproach yes, that's another way of putting it. $\endgroup$ – Joffan Feb 2 '17 at 17:43

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