Determine whether $\mathbb{Z}\times \mathbb{Z}$ and $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}$ are isomorphic groups or not. Determine whether $\mathbb{Z}\times \mathbb{Z}$ and $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}$ are isomorphic groups or not.
pf) Suppose that these are isomorphic. Note that $\mathbb{Z}\times \mathbb{Z}$ is a subgroup of $\mathbb{Z}\times \mathbb{Z}$ and $\mathbb{Z}\times \mathbb{Z}\times\left \{ 0 \right \}$ is a subgroup of $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}$. Since $\mathbb{Z}\times \mathbb{Z}$ and $\mathbb{Z}\times \mathbb{Z}\times\left \{ 0 \right \}$ are isomorphic, $\mathbb{Z}\times \mathbb{Z}/\mathbb{Z}\times \mathbb{Z}$ and $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}/\mathbb{Z}\times \mathbb{Z}\times \left \{ 0 \right \}$ are isomorphic. But the first one is isomorphic to the trivial group and the second one is isomorphic to $\mathbb{Z}$. It is a contradiction.
Is my proof right? If not, is there another proof?
 A: Your proof is not quite correct - an (abstract) homomorphism $\mathbb{Z}^2 \to \mathbb{Z}^3$ need not send $\mathbb{Z}^2$ to $\mathbb{Z}^2 \times \{0\}$. Here's my preferred way of showing they are not isomorphic (and the argument generalizes):
For any abelian group $A$, the set of group homomorphisms $\text{Hom}(\mathbb{Z}, A)$ has the same cardinality as $A$ (a bijection is given by $a \leftrightarrow (1 \mapsto a)$). Combining this with the fact that $\text{Hom}(\mathbb{Z}^n, A) \cong A^n$ gives that $\text{Hom}(\mathbb{Z}^2, A)$ and $\text{Hom}(\mathbb{Z}^3, A)$ have different cardinalities (if $1 < |A| < \infty$).
A: Suppose  $\mathbb{Z}\times \mathbb{Z}$ is isomorphic to $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}$, via a map $\phi$. Then, as $(1,0)$, $(0,1)$ generate  $\mathbb{Z}\times \mathbb{Z}$, $\phi(1,0)$ and $\phi(0,1)$ generate $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}$. But  $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}$ cannot be generated by fewer than 3 elements, a contradiction. So  $\mathbb{Z}\times \mathbb{Z}$ and $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}$ are not isomorphic.
A: We know that these two groups are free abelian in which for $\mathbb Z\oplus\mathbb Z$ and $\mathbb Z\oplus\mathbb Z\oplus\mathbb Z$ the basis sets don't have the same cardinal number , so according to this Theorem 3. or Theorem  10.14 the groups are not isomorphic. 
A: No, it is not correct. Suppose $f:\mathbb{Z}\times \mathbb{Z}\to\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}$ is an isomorophism. Then you have only $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}/\mathbb{Z}\times \mathbb{Z}\times \left \{ 0 \right \}\cong \mathbb{Z}\times \mathbb{Z}/f^{-1}(\mathbb{Z}\times \mathbb{Z}\times \left \{ 0 \right \})$.
A: The class of Abelian groups is an equational class. $\mathbb Z\times\mathbb Z$ is a free Abelian group with $2$ generators. $\mathbb Z\times\mathbb Z\times\mathbb Z$ is a free Abelian group with $3$ generators. If the free Abelian group with $2$ generators were isomorphic to the free Abelian group with $3$ generators, then it would follow from something in universal algebra [*] that there are no finite Abelian groups of order greater than $1$. But there is an Abelian group of order $2$. This contradiction proves that those two groups are not isomorphic.
[*] Namely, if $\mathbf K$ is an equational class (variety) of algebras (in the sense of universal algebra), and if the free $\mathbf K$-algebra on $m$ generators is isomorphic to the free $\mathbf K$-algebra on $n$ generators for some $m,n\in\mathbb N,m\ne n$, then $\mathbf K$ contains no finite algebra with more than one element. This is a basic theorem of universal algebra. Here is a proof for the case $m=2,n=3$:
Suppose $F$ is is a free $K$-algebra with free generating sets $\{a,b\}_\ne$ and $\{c,d,e\}_\ne$. Then there are "polynomials" (terms in the language of $\mathbb K$) $\varphi(x,y,z),\ \psi(x,y,z),\ f(u,v),\ g(u,v),\ h(u,v)$ such that $a=\varphi(c,d,e),\ b=\psi(c,d,e),\ c=f(a,b),\ d=g(a,b),\ e=h(a,b)$. Hence the following identities hold in every $K$-algebra:
$$u=\varphi(f(u,v),g(u,v),h(u,v))$$
$$v=\psi(f(u,v),g(u,v),h(u,v))$$
$$x=f(\varphi(x,y,z),\psi(x,y,z)$$
$$y=g(\varphi(x,y,z),\psi(x,y,z)$$
$$z=h(\varphi(x,y,z),\psi(x,y,z)$$
These identities show that, for any algebra $A\in\mathbf K$, the mapping
$$\langle u,v\rangle\to\langle f(u,v),g(u,v),h(u,v)\rangle$$
is a bijection from $A\times A$ to $A\times A\times A$. It follows that $|A|^2=|A|^3$, i.e., $|A|$ is infinite or $|A|=1$.
