# 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.

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.

• Thank you very much for your answer. You are completely right for the symmetric and dihedral group, that was a stupid question on my part. I think I understand better why we use the group $\mathbb{Q}/\mathbb{Z}$. Nevertheless, I don't understand how it gives a geometric/intuition about the orthogonal of a group. I already knew your analogy, but I don't understand how it helps understanding the orthogonal of a subgroup. Also I don't understand your last sentence : "... separates points of $A$". What separates points means and how did you conlude that ? (so there is an intuition about the dual?) – auhasard Dec 2 '18 at 18:46
• It doesn't really give geometric intuition, unless you already have some concerning the orthogonal of a subvector space. If you do, then you can think of it as being similar, so it does give some geometric insight. $\check{A}$ separates points of $A$ means if $x\neq y$, then there is some $f\in\check{A}$ such that $f(x) \neq f(y)$, and it follows directly from what I mentioned. – Max Dec 2 '18 at 18:53
• +1Thank you. The problem is that for me the orthogonal of a subvector space $W$ is:$\{x \in V \mid B(x,y) = 0 \forall y \in W\}$ where $B$ is a bilinear form hence it's composed of vectors not of functions whereas the orthogonal of a subgroup is a set of functions. Hence for me the orthogonal of a s.v.s $W$ is really just the space orthogonal to $W$ in $V$, so geometricaly this is very intuitive. For the orthogonal of a group it doesn't seem like I can relate the element of $B^\perp$ with the one of $B$ since there are different (one are morphisms whereas in $B$ we just have element of a g $G$ – auhasard Dec 2 '18 at 19:02
• That's a version of the orthogonal when you are given a bilinear form $B$. Similarly if you were given a bilinear map $A\times A\to \mathbb{Q/Z}$, you could define the orthogonal subgroup as a subgroup of $A$. But for a general vector space $V$, with no given bilinear form, the orthogonal is indeed a subvector space of $V^*$. To relate the two, note that if you are given a bilinear form $B$, you have a clear map $V\to V^*$ (analogously $A\to \check{A}$) defined by $j: x\mapsto (y\mapsto B(x,y))$, and then the orthogonal subvector space of $W\leq V$ is $j^{-1}(W^\bot)$, where $W^\bot\leq V^*$ – Max Dec 2 '18 at 19:10
• (and similarly $j^{-1}(B^\bot)$, where $B^\bot \leq \check{A}$). Here in some sense it's more general, it's a "universal" orthogonal because you can find any orthogonal defined by a bilinear form thanks to it – Max Dec 2 '18 at 19:11