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It's not clear for me how to find the appropriate transversal element in the next example below for an example for $x,y,z$ elements.

Please would someone show this process completely?

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Metso
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    There is a worked example of Reidemeister-Schreier in section III.8 of course notes of mine: https://www.math.colostate.edu/~hulpke/CGT/cgtnotes.pdf – ahulpke Aug 26 '19 at 15:39
  • @ahulpke Thank you – Metso Aug 26 '19 at 16:07
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    You might find this old answer of mine helpful. See also this worked example. (Both examples take a more topological point of view.) – user1729 Aug 27 '19 at 08:14
  • @user1729: thank you for your respond. The Reidemeister-Schreier algorithm could be used for normal and not normal subgoup (abnormal). Am I right? – Metso Sep 11 '19 at 05:15
  • @Metso yes, the algorithm works for all subgroups. It's particularly nice for normal ones though, as the cosets are easier to find. – user1729 Sep 11 '19 at 07:12
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    (Also, I've never heard non-normal subgroups called "abnormal" before. I would advise just calling them "subgroups". Giving them a name implies that they have a property, rather than lacking a property. For example, there is a type of subgroup called "malnormal", and people may assume abnormal=malnormal.) – user1729 Sep 11 '19 at 07:13
  • @user1729: ok, agree with your terminology. If a subgroup is not normal, is it enough to find left or right cosets for R-S method, but not both? – Metso Sep 11 '19 at 08:31
  • @Metso Yes, just find left cosets or right. You choose. – user1729 Sep 11 '19 at 10:43

1 Answers1

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The group $S_3$ is defined as $\langle a,b \mid a^2 = b^3 = 1, ba = ab^2\rangle$, where $a = (12), b = (123)$.

The elements $ba$ and $ab ^ 2$ are both equal $(23)$. Since there is a homomorphism from $G$ to $G/H = S_3$, the elements $ba$ and $ab ^ 2$ are in the same coset. Therefore, the Schreier representative $ba$ is equal to $ab ^ 2$.

Viktor Vaughn
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Metso
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