orthogonal group what does it represent 
Let $A$ be an finite abelian group and $B$ be a subgroup of $A$. Then we defined the orthogonal of $B$ : 
  $$B^{\perp} = \{f:(A,+) \to (\mathbb{Q}/\mathbb{Z},+) \mid \forall b \in B ,f(b) = 0 \}$$

I understand the definition, but I don't understand what it really represents. In linear algebra, the orthogonal of a sub-vector space has a very nice geometric meaning. Here I don't understand what it represents and why it's useful? 
Hence here I am really seeking for intuition. For example, if I take famous groups like the dihedral group of the symmetric group and take a subgroup of one of these group. Then what the orthogonal will represent? 
I we think of a group as a set of transformations that can act on some set $X$ maybe the orthogonal has a more geometric meaning? 
Finally, why do we take morphism that go to the group : $\mathbb{Q}/ \mathbb{Z}$ ? I know this has a link with representation theory but isn't there a simple explanation to take this group ? 
Thank you.
 A: Note that you can't really take the dihedral or symmetric groups, as this is about abelian groups. 
Here's an interesting analogy : for a vector space $V$, you have a dual vector space $V^*:= \hom(V,k)$, and if $W\leq V$ is a subvector space, then $W^\bot\leq V^*$ is a subvector space. Here, for an abelian group $A$ you have a dual abelian group $\check{A} := \hom (A, \mathbb{Q/Z})$, and given $B\leq A$ a subgroup you have $B^\bot \leq \check{A}$. 
Essentially it will be as useful as $V^*,W^\bot$ are in the context of vector spaces. 
The reason why $\mathbb{Q/Z}$ is that, just as $k$ is a cogenerator in the category of vector spaces, $\mathbb{Q/Z}$ is one in the category of abelian groups. What does it mean ? Here, essentially, it reduces to the fact that given $x\in A\setminus\{0\}$, there is a map $A\to \mathbb{Q/Z}$ such that $f(x) \neq 0$, this is because $\mathbb{Q/Z}$ is injective, i.e. given a subgroup $B\leq A$ and a map $f:B\to \mathbb{Q/Z}$, there is a map $g:A\to \mathbb{Q/Z}$ extending $f$. 
Then for $x\neq 0 \in A$, either $x$ has a finite order $n$, in which case $x\mapsto \frac{1}{n}$ mod $\mathbb{Z}$ gives a map $\langle x\rangle \to \mathbb{Q/Z}$ which sends $x$ to something nonzero; either $x$ has infinite order, in which case you can send it to anything nonzero in $\mathbb{Q/Z}$; in either case there is a nonzero $f:\langle x\rangle \to \mathbb{Q/Z}$, which can then be extended to $A$. 
So $\check{A}$ separates points of $A$, and so plays a similar role to $V^*$ in the context of $k$-vector spaces.
