The difference between $\mathbb{Z}\times\mathbb{Z}$ and $\mathbb{Z}*\mathbb{Z}$?

Question

I have some questions regarding some notations being used here.

I am still relatively new to algebraic topology, so I am a bit confused.

I saw that $\pi_1(S^1\times S^1)\simeq\mathbb{Z}\times\mathbb{Z}$ and $\pi_1(S^1\vee S^1)\simeq\mathbb{Z}*\mathbb{Z}$.

I know the difference between $\times$ and $\vee$. But what I am unsure is the difference between $\mathbb{Z}\times\mathbb{Z}$ and $\mathbb{Z}*\mathbb{Z}$.

$\mathbb{Z}\times\mathbb{Z}$ is the product group am I right? But what is $\mathbb{Z}*\mathbb{Z}$? How do we call it? I could not search since I don't know the name.

Could somebody please give some help? Thanks.

Answer 1

The notation $*$ denotes that you are considering the free product of the two groups. The free product of two groups is basically words in an alphabet given by them. However, we also require that each word is fully reduced (i.e $aa^{-1} = e, aa = a^2$) in the dictionary of words that can be described by the alphabet.

Answer 2

$\mathbb{Z}*\mathbb{Z}$ is the free product of two copies of the integers. The free product of $G$ and $H$ is created by taking all of the words $s_1s_2s_3\dots s_n$ where $s_i$ are elements of $G$ or $H$, and words may be reduced by removing the identity wherever it appears, and if two adjacent elements are part of the same group, you may replace them by their product within the group. One can see that every reduced word consists of alternaitng elements of the groups given. One thing that you will notice is that even if $G$ and $H$ are abelian, their free product need not me (and in fact, I believe it never is). As you learn about the Seifert Van Kampen theorem you will see that under good conditions the fundamental group functor preserves pushouts of objects, and the pushouts in the category of groups are free products of groups mod a normal subgroup represented by the maps in the pushout. Its usually quite difficult to work with these, as you get a free group on some number of elementsw and some number of relations, but often much progress can be made in describing the group. One thing you can often look at is the abelianization of the group, which corresponds to $H_1$.

Answer 3

$\mathbb{Z}\times\mathbb{Z}$ is just the Cartesian product, i.e., the set $\{(a,b)|a,b\in\mathbb{Z}\}$ with group operation.

$\mathbb{Z}*\mathbb{Z}$ is the free product. You should read about it in full generality$^1$, but if we let $a$ be the generator of the first $\mathbb{Z}$ and $b$ be the generator of the second $\mathbb{Z}$, then the free product $\mathbb{Z}*\mathbb{Z}$ is the set of elements of the form $g=a^{i_1}b^{j_1}\cdotp\cdotp\cdotp a^{i_m}b^{j_m}$ where $i_1$ or $j_m$ may be zero, but none of the other exponents are, along with the group operation of concatenation.

Do you see why this is the case? $a$ represents a loop going around one of the copies of $\mathbb{S}^1$ once, and $b$ represents a loop going around the other copy of $\mathbb{S}^1$ once. Intuitively, the set of loop classes of $\mathbb{S}^1\vee\mathbb{S}^1$ will be loops that go around one of the copies of $\mathbb{S}^1$ some integer number of times, then go around the other copy of $\mathbb{S}^1$ some integer number of times, and so on. This is exactly what $\mathbb{Z}*\mathbb{Z}$ encapsulates.

$^1$ I recommend Hatcher's Algebraic Topology or Lee's Introduction to Topological Manifolds.