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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 Klein-$4$ 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!

Shaun
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solo.
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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 06:32

1 Answers1

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HINT:

  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.

Brian M. Scott
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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 – user10024395 Nov 06 '14 at 05:57
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    @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 06 '14 at 07:57
  • 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