Question

What are two different groups of order 6 that are isomorphic?

Answer 1

$\huge\bold\pink{ANSWER }$

I need to show that there are only 2 groups of order 6 up to isomorphism.

I did prove it, but the proof is quite cumbersome. I wonder if there is a very concise proof.

My proof outline: Suppose G is of order 6 and is not Z6. Then any element that is not identity must have order 2 or 3.

I went on to show that there must be a element of order 2 and another one of order 3, and the intersection of the cyclic subgroups generated by the two is the identity. Then I'm able to show it is isomorphic to S3.

Answer 2

A group of order 6 is isomorphic to Z/(6) or to S3. 3 then xz has order 6 since (xz)6 = e, (xz)2 = x2z2 = z2 = e, and (xz)3 = x3z3 = x = e. Thus again G is cyclic, so G ∼ = Z/(6).