I know what a free group is, its intuition as well as its formal rigorus definition using the university property - namely that a free group of a set $S $ is the set (in fact a group) $F (S) $ such that for any map $f $ from $S $ to any group $G $ can be extended uniquely to a map $g: F (S) \to G$ such that $f = g \circ inc $ where $inc $ is the inclusion map from $S $ to $F (S) $.
I know that, intuitively, that $K := D_{2•4} $ (written this way as to avoid confusion; group isomorphic to rigid rotation a square, of order $8$) has a relation and cannot be a free group. But I am not sure how to rigorously prove that it is not a free group. Specifically, assuming $K = F (S) $ for a set $S $ (which is a subset of $K$), how may we find a group $G$ and a funtion $f $ such that for every map $g: F (S) \to G $ satisfy $f \neq g \circ inc $?
The point is that any set of generators of $K$ will have some relations between them, equations that are true of them but which do not need to be true in an arbitrary group. Then if you take $G$ to be a group with elements that do not satisfy those relations, you will get a counterexample to the universal property.
In particular, let $s\in S$ be any element. Since $K$ is finite, $s$ has finite order, so there is some $n>0$ such that $s^n=1$. But this is not true of every element in every group; in particular, it is not true of any nonzero element of $\mathbb{Z}$. So, let's take $f:S\to\mathbb{Z}$ to be any map such that $f(s)=1$. If we could extend $f$ to a group-homomorphism $g:K\to\mathbb{Z}$, we would have $g(s^n)=n\cdot g(s)=n$, which is a contradiction since $s^n=1$ and so we must have $g(s^n)=0$.
The proof that $K$ is not free is not quite complete yet, because I assumed that the set $S$ had an element! We must still rule out the possibility that $K$ is free on the empty set. In this case, actually, any map $f:\emptyset\to G$ always can be extended to a homomorphism $f:K\to G$, since we can take just the trivial homomorphism. However, the extension will not always be unique (for instance, if $G=K$, $g$ could be either the trivial homomorphism or the identity map).
This argument more generally shows that any nontrivial finite group is not free. (Nontriviality is required for the last part--in fact, the trivial group is free on the empty set.)