If #f# is a group homomorphism from #S_3# into #ZZ_6#, then what is the order of #f(S_3)# ?

1 Answer
Sep 9, 2017

#1# or #2#

Explanation:

Since #ZZ_6# contains an element of order #6#, but #S_3# does not, there is no group isomorphism. So the image of #S_3# in #Z_6# is of order strictly dividing #6#.

Consider the image of the transposition #((1, 2))# under #f#. It must either be an element of order #1# or of order #2#. It cannot have order #3#.

So we either have #f(((1,2))) = hat(0)# or #f(((1,2))) = hat(3)#

Note that if #((m,n)) in S_3# then there is some permutation #p# in #S_3# such that: #p^(-1) ((1, 2)) p = ((m, n))#. Hence:

  • If #f(((1,2))) = hat(0)# then #f(((m,n))) = hat(0)# for all transpositions in #S_3# and hence #f(S_3) = { hat(0) }#

  • If #f(((1, 2))) = hat(3)# then #f(((m,n))) = hat(3)# for all transpositions in #S_3# and hence #f(S_3) = { hat(0), hat(3) }#