Is one-to-one-ness and operation preservation sufficient to prove an isomorphism between two finite groups with identical cardinality?

Question

See question.

In my Abstract Algebra textbook, it is mentioned that to prove an isomorphism $\phi$ exists is to show it is well defined, one-to-one, onto, and operation preserving.

It seems like almost obvious to me, since a one-to-one function between sets A and B with |A| = |B| is necessarily onto, that if we know the groups are of identical order, then wouldn’t it be enough to just prove one-to-one-ness and operation preservation?

Sorry - I’m new to this and since my textbook hasn’t ever used this (seemingly easy?) method, Im suspicious.

Thanks

Answer 1

If the groups have the same finite cardinality, yes, you just need to check that you have an injective homomorphism to see that you have an isomorphism.

If the codomain has greater cardinality, then you still get an isomorphism onto the image.

That's in either case you automatically get an inverse homomorphism from the range.

So it's a handy simplification.