How to prove that $\mathbb Z\times \mathbb Z$ is not free?

Question

We knew that the additive group $\mathbb Z\times \mathbb Z$ is an abelian but non-cyclic group.

Why $\mathbb Z\times \mathbb Z$ is not free? But it is a free abelian group with the basis $S = \{(1; 0), (0; 1)\}.$

Thanks all for help!

Answer 1

Given any two distinct element $a,b\in\mathbb{Z}\times\mathbb{Z}$, there is no homomorphism $f\colon\mathbb{Z}\times\mathbb{Z}\to S_3$ such that $$ f(a)=(12)\qquad f(b)=(13) $$ because we'd have $$ (132)=(12)(13)=f(a)f(b)=f(a+b)=f(b+a)=f(b)f(a)=(13)(12)=(123) $$ which is a contradiction.

Therefore $\mathbb{Z}\times\mathbb{Z}$ cannot be free over any basis with at least two elements; on the other hand it is not a free group over one element, because $\mathbb{Z}\times\mathbb{Z}$ is not isomorphic to $\mathbb{Z}$, not being cyclic.

More generally, no two distinct elements in a basis of a free group commute, so no free group other than (up to isomorphism) $\mathbb{Z}$ is abelian.

Answer 2

It is a free abelian group, but not a free group.

Answer 3

It is not free because $a+b=b+a$ for all $a,b$. In a free group, for two independent generators $a$ and $b$ we have $ab\ne ba$.

Note that it is a free abelian group, but not an abelian free group. The word order matters.

Answer 4

To understand this, it may help to know a little category theory...

It depends which category you are working in... it's free in the category of abelian groups, but not in the category of groups. ..

"Free" is a categorical concept, namely that of satisfying a certain universal property...

It just so happens to also mean being "free" of any relations...

$a+b=b+a $ is a relation in the category of groups; but not in that of abelian groups. ..