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Can the following statement be proved? "there are only two different groups of order 4 up to isomorphism" I have seen somewhere that there are only two groups up to isomorphism of order 4 -cyclic of order 4 and the klein4 group.All other groups with 4 elements are isomorphic to one of these.Really I don't have any idea on attempting this problem, so I please need your help.Thanks all!

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Recall that the order of an element always divides the size of the group. Only one element can have order $1$ (the identity). What options are left? – Andrés E. Caicedo Jan 21 '12 at 6:32


  1. What happens if the group has an element of order $4$?

  2. If it has no such element, the three non-identity elements all have order ... what? Then give them names $-$ $a,b$, and $c$, for instance $-$ and start filling out the group multiplication table. You’ll find that there’s only one way to do so that’s consistent with the properties required of a group.

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Just wondering, for (2), is there a more effective ways of knowing the number of elements of each order, here order 1 and 2, other than filling in the multiplication table. Because if the order of the group is bigger, filling in multiplication table is not feasible – user136266 Nov 6 '14 at 5:57
@user136266: In general that’s not an easy question, but neither is the question of how many non-isomorphic groups there are of a given order. – Brian M. Scott Nov 6 '14 at 7:57
I don’t suppose that the downvoter would care to be useful and explain what’s wrong? No, I didn’t think so. – Brian M. Scott Apr 14 '15 at 20:55

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