Do group algebras of the reals over finite groups contain anything other than copies of $\mathbb{C}$ and $\mathbb{R}$?

Sorry if the question is ambiguous/non-formal, I'm happy to accept any edits that will clarify.

I'll try to set it up better:

Let $\mathfrak{A} = \mathbb{R}[\mathbb{Z}_n]$ be a group algebra of the reals over the integers modulo $n$ (with addition as the group action). I'm about 90% sure that my term is correct, but what I mean is this: let $\{r_1,r_2,...,r_n\}$ be a set of real numbers. The $v \in \mathfrak{A}$ means that $v = r_1a_1 + r_2a_2 +...+r_na_n$ where $a_i$ is the $i^{th}$ element of $\mathbb{Z}_n$, where addition is standard vector addition and multiplication is a mixture of pure real multiplication and the group action. Again, my apologies if this is the wrong term/too ambiguous.

Since $\mathbb{Z}_n$ under addition is isomorphic to the group formed by the $n^{th}$ roots of unity under multiplication, this group algebra can be thought of as a vector space over the complex numbers.

I'm interested in exploring what happens when the group under consideration, $G$, is not $\mathbb{Z}_n$. I've already come to a few conclusions:

1. Let $g \in G$ be an element of $G$ such that $g = g^{-1}$ under the group operation. Then, since $g$ generates a subgroup isomorphic to $\mathbb{Z}_2$. Any element of the algebra $\mathbb{R}[G]$ whose coefficients are zero everywhere except at the index corresponding to $g$ is the same as (isomorphic to?) $\mathbb{R}$.
2. Let $g \in G$ be an element of $G$ such that $g \ne g^{-1}$. Then $g$ generates a subgroup isomorphic to $\mathbb{Z}_m$ for some $m$ less than the order of $G$. A similar construction as above says that the elements of the algebra $\mathbb{R}[G]$ whose coefficients are zero except at the indices corresponding to the the subgroup generated by $g$ will be isomorphic to the algebra $\mathbb{R}[\mathbb{Z}_m]$, and can be thought of as a vector space over the complex numbers.

Now here's where I'm stuck:

$G$ may contain other sub-groups that are not isomorphic to $\mathbb{Z}_m$ for some $m$. However, what about these sub-groups? Is it possible for a group to contain no sub-groups isomorphic to $\mathbb{Z}_m$ (at least for $m \gt 2$?) If so, how would my interpretation of the group algebra change?

• I believe the convention is to call $\mathbb{R}[\mathbb{Z}_n]$ the group algebra of $\mathbb{Z}_n$ over $\mathbb{R}$ (in general, $K[G]$ (or $KG$) the group algebra of $G$ over $K$). Anyway, I would like to know how you make $\mathbb{R}[\mathbb{Z}_n]$ into a complex vector space, because I see no canonical way to do that (one which restricts to the obvious real vector space structure). Moreover, your first observation seems to be not correct to me, for the set of elements of the algebra $\mathbb{R}[G]$ whose coefficients are zero everywhere except at $g$ is not closed under multiplication. – Matthias Klupsch Jun 28 '17 at 12:31
• I may be using the wrong term. Consider two elements of $\mathbb{R}[\mathbb{Z}_3]$: $v = r_1a + r_2b + r_3c$, $w = r_4a + r_5b + r_6c$, then $vw = r_1r_4a\circ a + r_1r_5a\circ b + r_1r_6a\circ c +...$ is my multiplication operation, an $v+w = r_1+r_4 a + r_2 + r_5 b + r_3 + r_6 c$ is my addition. If I am not mistaken this does conform to the required axioms of vector spaces (eventually, too hard to show in a comment). We might be amused to plot this with $a = \langle 1, 0,0\rangle$, $b = \langle 0, 1, 0 \rangle$ etc. and that might show geometrically that we do in fact get a vector space. – Michael Stachowsky Jun 28 '17 at 13:45
• I don't have enough room in the comments to give my full definition, but does the structure I'm talking about make sense now, given the extra description? – Michael Stachowsky Jun 28 '17 at 13:47
• For a scalar multiplication for the real numbers, yes, however, you want to have a complex vector space, i.e., a scalar mulitplication $\Bbb{C} \times \Bbb{R}[\Bbb{Z}_n] \to \Bbb{R}[\Bbb{Z}_n]$ and I claim there is no canonical one which extends the scalar multiplication from the reals. – Matthias Klupsch Jun 29 '17 at 4:09
• Still, you seem to be interested in the addition as well as the multiplication defined on the group algebra, right? So, it seems to me, you are looking for subrings and subalgebras of $\mathbb{R}[\mathbb{Z}_n]$. – Matthias Klupsch Jun 29 '17 at 12:52

Concerning the subgroup question, it is quite possible that $G$ does not contain any subgroup isomorphic to $C_m$ for $m>2$. For example, take $G=C_2\times C_2$, or $G=C_2^n$ for some $n\ge 2$. But I don't see a problem then for the group ring or group algebra $k[G]$, which you can consider for any field $k$ and any group $G$.
• Well, what is your problem with, say, $C_2\times C_2$? – Dietrich Burde Jun 28 '17 at 13:54