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Observe that their centers are not isomorphic.
Z(Sn) = {e}, while {0} x Z/2Z is in the Z(An x Z/2Z).
(We can be more precise; Z(A3) = A3, since A3 is abelian, while Z(An) = {e] for n > 3.
==> Z(A3 x Z/2Z) = Z/3Z x Z/2Z ≅ Z6, and Z(An x Z/2Z) = {e} x Z/2Z ≅ Z/2Z for n > 3.)
I hope this helps!
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Observe that their centers are not isomorphic.
Z(Sn) = {e}, while {0} x Z/2Z is in the Z(An x Z/2Z).
(We can be more precise; Z(A3) = A3, since A3 is abelian, while Z(An) = {e] for n > 3.
==> Z(A3 x Z/2Z) = Z/3Z x Z/2Z ≅ Z6, and Z(An x Z/2Z) = {e} x Z/2Z ≅ Z/2Z for n > 3.)
I hope this helps!