[Math] Prove that there only two groups of order 4 up to isomorphism

abstract-algebrafinite-groupsgroup-isomorphismgroup-theory

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!

Best Answer

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.

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