Let $\mathbb{Z}^n$ be the group formed by the external direct product of $\mathbb{Z}$ taken $n$ times, and let $A_n$ be the group formed by taking the free product of $\mathbb{Z}$ $n$ times.
Then, is $\mathbb{Z}^n \cong A_n $? I was told in class that both of these groups have the same presentation i.e. generated by a set of $n$ elements, with only the trivial relation i.e. both of them are isomorphic to $F_n$, where $F_n$ is the free group with rank n. However, I'm confused as to how they can be isomorphic, as $\mathbb{Z}^n$ is abelian and $A_n$ is not.
Edit: If they are not isomorphic, then what is the presentation for each of these groups?
Best Answer
$\mathbb{Z}^n$ is typically defined as the direct sum of $n$ copies of $\mathbb{Z}$. Also known as the free abelian group.
On the other hand, the free product of $n$ copies of $\mathbb{Z}$, which you denote by $A_n$, is isomorphic to the free group $F_n$ (note: without "abelian") on $n$ generators.
These groups are isomorphic only for $n=1$. When $n>1$ then $F_n$ is not abelian while $\mathbb{Z}^n$ is. Formally presentations are as follows:
$$F_n=\big\langle \{g_1,\ldots, g_n\}\ \big|\ \emptyset \big\rangle$$ $$\mathbb{Z}^n=\big\langle \{g_1,\ldots, g_n\}\ \big|\ \{g_ig_kg_i^{-1}g_k^{-1}\}_{i,k=1}^n\big\rangle$$
Or a bit less formally:
$$F_n=\big\langle g_1,\ldots, g_n\big\rangle$$ $$\mathbb{Z}^n=\big\langle g_1,\ldots, g_n\ \big|\ g_ig_k=g_kg_i\text{ for any }i,k\big\rangle$$
So as you can see they don't have the same presentation: $\mathbb{Z}^n$ has additional "commutativity" relations. Besides they can't have the same presentation, because the same presentation implies isomorphism. Unless someone approaches the topic in a non-standard way (e.g. ignoring commutativity relations), but then he/she should be explicit about it. It is very unlikely for a typical mathematician to think that $\mathbb{Z}^n$ and $F_n$ have the same presentation.