Definition of dual group. The following definition is given in Dummit and Foote's Abstract Algebra on page 146.
For any group $G$ define the dual group of $G$ (denoted $\hat{G}$) to be the set of all homomorphisms from $G$ into the multiplicitive group of roots of unity in $\mathbb{C}$. Define a group operation in $\hat{G}$ by point wise multiplication of functions: if $\chi$ and $\psi$ are homomorphisms from $G$ into the group of roots of unit, then $\chi \psi$ is the homomorphism given by $(\chi \psi)(g)=\chi(g)\psi(g)$ for all $g\in G$, where the latter multiplication takes place in $\mathbb{C}$. 
I am confused as to what is meant here by "the multiplicitive group of roots of unity". I have only studied the multiplicitive group of $n$th roots of unity. Is it the group of $n$th roots for every $n$? Thanks. 
 A: The multiplicative group of $n$th roots of unity, which I'll call $\mu_n$, is the set of complex numbers $z$ such that $z^n=1$. Geometrically, $\mu_n$ is the set of points on the unit circle in $\mathbb C$ lying at angles $2\pi k/n$, $k=0,1,\dots,n-1$. $\mu_n$ is basically the same as the additive group $\mathbb Z/n\mathbb Z$: these groups are isomorphic.
But $\mu_n$ is not the group that Dummit and Foote are talking about. Your problem concerns the full multiplicative group of the roots of unity, which I'll call $\mu$. This consists of all complex numbers that are a root of $+1$. Formally,
$$
\mu = \bigcup_{n=1}^\infty \mu_n
$$
For instance, $1,-1,i,(1+i)/\sqrt2\in\mu$, and more generally, $e^{2\pi i k/n}\in\mu$
for any $k,n\in\mathbb Z$. 
Geometrically, $\mu$ is the set of points on the unit circle in $\mathbb C$ lying at angles that are rational multiples of $2\pi$.
But unlike $\mu_n$, $\mu$ is not isomorphic to any group of the form $\mathbb Z/n\mathbb Z$. Instead, $\mu$ is isomorphic to the group $\mathbb Q/\mathbb Z$: the rational numbers modulo $1$.
A: It's the multiplicative subgroup of $\mathbb{C}^*$
$$T=\{ z \in \mathbb{C} : \exists n \geq 1 | z^n=1\}$$
