Existence of nontrivial unit in $\mathbb{Q}[G]$, where $G$ is finite.

Question

Suppose $G$ is a finite group of order $|G|>1$, and $\mathbb{Q}[G]$ is the group ring. I'm curious about an example of a nontrivial invertible element, i.e., one that is not of the form $ag$, with $a\in\mathbb{Q}$ and $g\in G$.

This comes from an old UCLA qual problem 2002. An example is given at the bottom of this document on page 2, but it uses some facts about circulant matrices. I'm just curious if there is an alternative example, or just a general existence argument.

Answer 1

Since every finite group contains a cyclic subgroup of prime order, your question reduces to that, where the answer can be gotten by simple arithmetic with roots of unity (which probably reduces to the circulant matrix thing you are alluding to). However, in a slightly more general setting, Higman (in this paper) computes structure of the group of units of a group ring of a finite abelian group. (in general, this question is extremely hard).

Answer 2

(This is just a more in-depth version of Igor Rivin's answer in case you are new to these types of arguments).

For the cyclic group of order $n$, your group ring is isomorphic to $\mathbb{Q}[T]/(T^n - 1)$ which is just the direct sum of the cyclotomic fields $\mathbb{Q(\zeta_d)}$ for $d | n$ (including $d = 1$, so for instance if $n = p$ is prime you get $\mathbb{Q}(\zeta_p) \oplus \mathbb{Q}$... I may need to be a little more careful if $n$ is even, so let's just assume it's not).

To write down units in this, just pick something nonzero in each summand. For a concrete example, let's let $n = 3$ and write $\omega$ for $\zeta_3$. We have

$$ (a + b\omega_3, c) \in \mathbb{Q}(\omega) \oplus \mathbb{Q} $$ which will be a unit unless $c = 0$ and $a = b = 0$. Let's move them back to the group ring to verify that (some of them) are not "trivial" in your sense.

Write $G = \langle g \rangle$. The isomorphism goes (the first $=$ is just $g\mapsto T$, the first arrow is the chinese remainder theorem, and the last evaluates $T$ to $\omega$ resp. $1$)

$$ \mathbb{Q}[G] = \mathbb{Q}[T]/(T^3 - 1) \to \mathbb{Q}[T]/(T^2 + T +1) \oplus \mathbb{Q}[T]/(T-1) \to \mathbb{Q}(\omega) \oplus \mathbb{Q} $$

We want to go backwards along these identifications, so we need to unpack the Chinese remainder theorem.

We seek $\alpha, \beta \in \mathbb{Q}[T]/(T^3-1)$ such that $\alpha + \beta = 1$, $(T^2 + T + 1)|\alpha$, and $(T - 1) | \beta$. Then the preimage of $(f, g)$ under the Chinese remainder theorem map will be $\beta f + \alpha g$.

Using the Euclidean algorithm (or guessing and checking) gives $\alpha = \frac{1}{3}(T^2 + T + 1)$ and $\beta = \frac{-1}{3}(T-1)(T+2)$.

We are done! $(a + b\omega_3, c)$ lifts to

$$ \frac{1}{3}\left(-(g-1)(g-2)(a + bg) + (g^2 + g + 1)(c) \right). $$

I'm out of time, so I'll leave it to you to tidy this expression up and see when it is non-trivial (and check to make sure I didn't make any mistakes!) Note that $g^3 = 1$ simplifies things.