Let D be the diagonal subset \(\displaystyle{D}={\left\lbrace{\left({x},{x}\right)}{\mid}{x}∈{S}{3}\right\rbrace}\) of the direct product S3xS3, where

\(\displaystyle{S}{3}={\left\lbrace{e},{\left({1},{2}\right)},{\left({2},{3}\right)},{\left({1},{3}\right)},{\left({1},{2},{3}\right)},{\left({1},{3},{2}\right)}\right\rbrace}\)

To show that D is a subgroup of S3xS3. We know that \(\displaystyle{\left({e},{e}\right)}∈{D}\) so, D is non-empty. Now suppose \(\displaystyle{\left({x},{x}\right)},{\left({y},{y}\right)}∈{D}\) for some \(\displaystyle{x},{y}∈{S}{3}\). Then

\(\displaystyle{\left({x},{x}\right)}\cdot{\left({y},{y}\right)}^{{-{{1}}}}={\left({x},{x}\right)}{\left({y}^{{-{{1}}}},{y}^{{-{{1}}}}\right)}={\left({x}{y}^{{-{{1}}}},{x}{y}^{{-{{1}}}}\right)}∈{D}.\)

Therefore, D is subgroup of S3xS3. Let \(\displaystyle{g},{h}∈{S}{3}{x}{S}{3}\), such that g=((1,2),(1,3)) and \(\displaystyle{h}={\left(\begin{array}{cc} {1}&{2}\\{1}&{2}\end{array}\right)}∈{D}\)

Now PSKg^-1=((1,2),(1,3)).((1,2),(1,2)).((1,2),(1,3))^-1 =((1,2),(1,3)).((1,2),(1,2)).((1,2)^-1,(1,3)^-1) =((1,2),(1,3)).((1,2),(1,2)).((1,2),(1,3)) =((1,2),(1,2).(1,2),(1,3).(1,2),(1,3).) =((1,2),(1,3).(1,3,2)) =((1,2),(2,3))∈ D.ZSK

Therefore, D is not normal subgroup of S3xS3.

\(\displaystyle{S}{3}={\left\lbrace{e},{\left({1},{2}\right)},{\left({2},{3}\right)},{\left({1},{3}\right)},{\left({1},{2},{3}\right)},{\left({1},{3},{2}\right)}\right\rbrace}\)

To show that D is a subgroup of S3xS3. We know that \(\displaystyle{\left({e},{e}\right)}∈{D}\) so, D is non-empty. Now suppose \(\displaystyle{\left({x},{x}\right)},{\left({y},{y}\right)}∈{D}\) for some \(\displaystyle{x},{y}∈{S}{3}\). Then

\(\displaystyle{\left({x},{x}\right)}\cdot{\left({y},{y}\right)}^{{-{{1}}}}={\left({x},{x}\right)}{\left({y}^{{-{{1}}}},{y}^{{-{{1}}}}\right)}={\left({x}{y}^{{-{{1}}}},{x}{y}^{{-{{1}}}}\right)}∈{D}.\)

Therefore, D is subgroup of S3xS3. Let \(\displaystyle{g},{h}∈{S}{3}{x}{S}{3}\), such that g=((1,2),(1,3)) and \(\displaystyle{h}={\left(\begin{array}{cc} {1}&{2}\\{1}&{2}\end{array}\right)}∈{D}\)

Now PSKg^-1=((1,2),(1,3)).((1,2),(1,2)).((1,2),(1,3))^-1 =((1,2),(1,3)).((1,2),(1,2)).((1,2)^-1,(1,3)^-1) =((1,2),(1,3)).((1,2),(1,2)).((1,2),(1,3)) =((1,2),(1,2).(1,2),(1,3).(1,2),(1,3).) =((1,2),(1,3).(1,3,2)) =((1,2),(2,3))∈ D.ZSK

Therefore, D is not normal subgroup of S3xS3.