Let G:=<a,b,c |a^2=b^2=c^2=(ab)^3=(ac)^2=(bc)^3=e> a) Consider the element w=aba

Question

Let G:=<a,b,c |a^2=b^2=c^2=(ab)^3=(ac)^2=(bc)^3=e>
a) Consider the element w=abacba of G, so w can be viewed as a word with 6 letters. Prove that if we add a letter a,b,or c to the beginning or end of w, then when we apply appropriate relations from G we can reduce to a word that is shorter than 6 letters.
b) The empty word e has no letters, and has length zero. Assume that there are exactly 3 distinct words of length 1, exactly 5 distinct words of length 2, exactly 6 distinct words of length 3, exactly 5 distinct words of length 4, exactly 3 distinct words of length 5, and exactly 1 distinct word of length 6, for a total of 24 distinct words. Write down all 24 words.
c) With s1=(12), s2=(23), and s3=(34), the symmetric group Sr=<s1,s2,s3>. We have s1^2=s2^2=s3^3=(s1s2)^3=(s1s3)^2=(s2s3)^3=e. prove G=~S4

Explanation / Answer

G:=<a,b,c |a^2=b^2=c^2=(ab)^3=(ac)^2=(bc)^3=e>

(1) Consider the word w =abacba.

                                aw = aa bacba

                                     = bacba (5 letters)

                               bw = babacba

Now (ab)^3=ababab =e

              so    baba = a-1b-1 =ab

So                          bw = abcba.(5 letters)

(2) Words of length 1 .........> a,b,c

Words of length 2--------------> ab,bc (ac=ca), ba, ca

Words of length 3..................> abc, bca, acb, bac, cba,cbc

Words of length 4.................> abca, abab,bcab,bcac,acbc

Words of length 5................> abcab,abcac,abcba

Word of length 6.....................> abcabc

(3) Note: There is a typo in the description of S4. s3 *2=e (not s3^2, as given)

The group G and the symmetric group S4 have identical (upto labelling) sets of generators and relations :

So mapping a to (12) , b to (23) and c to (34) provides an isomorphism.

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