The set of equivalence classes of closed paths in a space $V$ forms a group called the first homotopy group, denoted by $\pi_1(V)$. For example, there exist infinitely many different equivalence classes of closed paths in $\mathbb{S}^1$ each labelled by an integer $n\in \mathbb{Z}$. The first homotopy group, in this case, is $\pi_1(\mathbb{S}^1)=\mathbb{Z}$.
How do I show that different equivalence classes of paths form a group? In the above example, how do I prove that the group is $\mathbb{Z}$?
I have an physics background (that too, an elementary one) and have very little understanding of homotopy theory. I'm not sure what this means to say that different paths form a group.
For something to be a group it needs a set and a binary operation so first one must define a binary operation on (path-)homotopy classes of paths.
Suppose $[\gamma], [\delta]$ are equivalence classes and $\gamma : a \rightsquigarrow b, \,\delta:b\rightsquigarrow c$ are paths. Then define $[\gamma]\ast[\delta]=[\gamma \cdot\delta]$ to be our binary operation (you may wish to simplify and set $a=b=c$), where $\gamma\cdot\delta$ denotes the composition (concatenation) of paths. For this operation to be well-defined one must show that if $[\gamma]=[\gamma']$ and $[\delta]=[\delta']$ then $[\gamma\cdot\delta]=[\gamma'\cdot\delta'].$ In other words, you must take two path homotopies for $\gamma$ and $\delta$ and construct a third for $\gamma\cdot\delta.$
Recall now the rest of the definition of a group: we need our binary operation to be associative and to have an identity and inverse.
For associativity one needs to show that $[(\gamma_1\cdot\gamma_2)\cdot\gamma_3]=[\gamma_1\cdot(\gamma_2\cdot\gamma_3)]$ and this can be done by constructing a homotopy.
For identity use the empty loop which I shall call 1. For inverse’s just show that $\gamma\cdot\bar\gamma$ is homotopic to $1$ and then define $[\gamma]^{-1}=[\bar\gamma]$, where $\bar\gamma$ denotes the reversal of $\gamma.$ One May also wish to prove that the operation is well defined but this is very straightforward.
This would prove that homotopy classes of loops form a group under the given operations. (You May also want to prove closure, ie that if $\gamma,\delta$ are loops then so is $\gamma\dot\delta$ but this is almost by definition)
The next part is how one can find the group associated to a space and this is hard. Indeed much of the early part of a first course in algebraic topology deals with this question. The gist is like this:
Consider the map $p : \Bbb R\to S^1$ where $p(x)=\mathrm e ^{\mathrm i x}$. This is clearly continuous. Now take the point $1\in S^1$ and see that it’s fibre under $p$ is $\{0,\pm 2\pi,\pm 4\pi,\dots\},$ or in other words $2\pi\Bbb Z\cong \Bbb Z$ (where both are seen as groups under addition). Next one must show that any path $\gamma:[0,1]\to\Bbb S^1$ may be lifted to a path $\widetilde\gamma:[0,1]\to\Bbb R$ such that $\gamma = p\circ\widetilde\gamma,$ and that this path is unique for a given choice of $\widetilde\gamma(0).$ One can also prove that this works for homotopy classes of paths being lifted too. There is a step to observe that homotopies can be lowered through $p$ and then one can take any loop at 1 in $S^1$ and lift it to a path in $\Bbb R$ from 0 to $2k\pi$ for some integer $k$.
It turns out that this is a way to associate any integer to a path and because of the observed properties, each integer corresponds to exactly one equivalence class of loops through 1 in $S^1.$ The final step is to show that the group given for $S^1$ corresponds to integer addition, i.e the given correspondence is an isomorphism of groups. And then we are done.