How do you map a group-algebra into another algebra? Given a group with elements 1, -1 with the obvious group multiplication rules, I can create a group-algebra $\mathbb{R}[\mathbf{1},\mathbf{-1}]$. But this is a $2$ dimensional $\mathbb R$  algebra. What I would like is to map this algebra back into the real numbers.
Similarly given a group with elements 1,-1, i, -i. The group-algebra created from this is 4 dimensional over $\mathbb{R}$, whereas I want to map this into the complex numbers $\mathbb{C}$ or a pair of real numbers $\mathbb{R}^2$.
Another example is the group $\mathbb{Z}_3$ with elements 1,  $\omega$, $\omega^2$. The group algebra is $3$ dimensional over $\mathbb R$, but again I would like to map this onto the algebra of complex numbers with $\omega\rightarrow \frac{1}{2}(1+\sqrt{-3})$.
How do we express this mapping? In the first case it's because we want an equivalence between $a\mathbf{1}+b(\mathbf{-1})$ and $(a+c)\mathbf{1}+(b+c)(\mathbf{-1})$. So would we have to divide the algebra somehow by this equivalence? But still make sure we have a valid algebra.
 A: There isn't anything interesting about the group algebra itself that complicates things. You can determine what isomorphisms are possible by analyzing the structure.  They are all commutative semsimple algebras over $\mathbb R$, so they can only be made up of blocks of $\mathbb R$ and $\mathbb C$.
$\mathbb R[\{\pm 1\}]\cong\mathbb R^2$ which has four ideals, two of which allow mapping back onto $\mathbb R$.  The most attractive is probably the augmentation mapping $\sum k_gg\mapsto \sum k_g$. You will have this for every group algebra over $\mathbb R$, by the way.
$\mathbb R[\{\pm 1, \pm i\}]\cong \mathbb R[x]/(x^4-1)\cong\mathbb C\times\mathbb R\times \mathbb R$ has $8$ ideals, one allowing a mapping back onto $\mathbb R\times \mathbb R$, and one allowing a map back onto $\mathbb C$, and two allowing mappings back onto $\mathbb R$.
$\mathbb R[\{1,\omega,\omega^2\}]\cong\mathbb R[x]/(x^3-1) \cong \mathbb C \times \mathbb R$, so it admits one homomorphism onto $\mathbb C$ by modding out by the ideal generated by $1+\omega+\omega^2$ when you use precisely the map you described ($\omega$ mapping to a third root of unity in $\mathbb C$.)

Well what I would say is interesting about the group-algebra, is that you can take certain elements from ℝ such as 1 and -1 and form a group. Then the idea is to get back to the algebra you started with.

Of course, if you picked a subgroup $H$ of the multiplicative group of $\mathbb C$, there would be a unique $\mathbb R$-algebra homomorphism of $\mathbb R[H]\to \mathbb C$. That's guaranteed by the universal property of the group ring. For the cyclic group of order $n>2$, this is always going to produce a mapping back onto $\mathbb C$, because the image contains something on the real axis (namely $1$) and something complex (one of the complex roots of $x^n-1$) so that its image has $\mathbb R$ dimension $2$.
