Find all the distinct left cosects of H in G and hence index of H.

G be a group of all permutations of degree 3 on 3 symbol 1,2, &3 order of G= $ \mid G \mid$ = 3!=6

Give H={1,(1 2)} is a subgroup of G.

$\therefore order \ of\ H\ = \mid H \mid = \ 2!\ = 2$

by lagrange's theorem,

Index= (Number of different left cosets of subgroup H)

= $\frac {\mid G \mid}{\mid H \mid}$ = = $\frac {6}{2}$ = 3

Consider, Left coset of

(13H) = (13) {I, (1,2)}

= {(13)I, (13) (1,2)}

Now (13(12))= (1 2 3)

(13)H= {(13), (1 2 3)}

Similarly

(23)H= (23), {I,(1 2) 3}

= {(23)I, (23) (12)}

Now

(23) (12)= (132)

(23)H={(23),(132)}

Hence, the three distinct cosets of H are H, (13)H,(23)H

Index = 3