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Let $G$ be a finite solvable group of order $n$, and let $g_1 ... g_n$ be an enumeration of its elements. Let $a_1 ... a_n$ be a sequence of integers, such that $\sum a_i$ is relatively prime to $n$.

Consider $\mathbb{C}[G]$, the group ring of $G$ with complex coefficients. Does the element $\sum a_i g_i$ necessarily have to be a unit in the group ring? (I believe that the element does have to be a unit, and have a proof in the cyclic and abelian case, but was hoping for a reference in greater generality, at least in the case when $G$ is solvable.)

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This is false for the cyclic group of order $6$. Let $g$ be a generator. Then $g^2-g+1$ acts by $0$ on the representations where $g$ acts by a primitive $6$-th root of $1$, and hence is not a unit in the group ring, but $1-1+1=1$ is relatively prime to $6$.

Generalizing this example, the statement is false for the cyclic group of order $pq$, with $p$ and $q$ two different primes. Let $g$ generate this group and let $\chi: G \to \mathbb{C}^{\ast}$ be a character with $\chi(g)$ a primitive $pq$-th root of unity. Let $\Phi_{pq}(x) = \sum c_k x^k$ be the $pq$-th cyclotomic polynomial. So the element $\Phi_{pq}(g)= \sum c_k g^k$ in $\mathbb{Z}[G]$ acts by $\Phi_{pq}(\chi(g)) =0$ on the representation $\chi$. Thus $\sum c_k g^k$ is not a unit. On the other hand, $\sum c_k = \Phi_{pq}(1)= 1$. (To compute the last, note that $\Phi_{pq}(x) = \frac{(x^{pq}-1)(x-1)}{(x^q-1)(x^p-1)}$ and take the limit as $x \to 1$.)

I claim further that, if $G$ has any element of non-prime-power order, then $G$ fails to have this condition. Let $g$ be an element of order $pq$ and let $\chi: \langle g \rangle \to \mathbb{C}^{\ast}$ be an injective character. Let $V = \mathrm{Ind}_{\langle g \rangle}^G \chi$. Then $V$ restricted to $\langle g \rangle$ has $\chi$ as a summand, and this summand is in the kernel of $\Phi_{pq}(g)$ acting on $V$. So $\Phi_{pq}(g)$ acts non-injectively on a representation of $G$, and thus is not a unit.

So the only groups for which this might be right are groups where every element has prime power order. These were classified by Higman, so you can dig into his paper if you care enough.


On the positive side, the statement is true whenever $G$ is a $p$-group. Let $\alpha = \sum c_g g \in \mathbb{Z}[G]$. I will show that the determinant of $\alpha$ acting on $\mathbb{Z}[G]$ is $\left( \sum c_g \right)^{|G|} \bmod p$, and hence is not $0$ if $\sum c_g \not \equiv 0 \bmod p$. Reducing $\mathbb{Z}[G]$ modulo $p$, we get an action of $\alpha$ on $\mathbb{F}_p[G]$. More generally, I claim that $\alpha$ acts on any $G$-representation $V$ over $\mathbb{F}_p$ by $\left( \sum c_g \right)^{\dim V}$. This is simple: $V$ has a filtration whose associated graded is a $\dim V$-dimensional trivial representation. Passing to the associated graded doesn't change determinant, and $\alpha$ acts on the $1$-dimensional trivial representation by $\sum c_g$.

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    $\begingroup$ Alternatively, the product of this element with $(g + 1)(g^3 - 1)$ is $0$. (That's probably effectively the same proof in a minor disguise.) $\endgroup$ – LSpice Dec 11 '18 at 2:58
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    $\begingroup$ Maybe you’re missing some conditions in the edit? $S_3^{ab}\cong C_2$, whose order is not divisible by 3. $\endgroup$ – Jeremy Rickard Dec 11 '18 at 17:35
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    $\begingroup$ Thank you for the correction! Indeed, $S_3$ does seem to obey this condition. $\endgroup$ – David E Speyer Dec 11 '18 at 17:42
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    $\begingroup$ You still lose. $(x^2-x+1) + (x^5+x^4+x^3+x^2+x+1)$ has nonnegative coefficients and acts by $0$ on the same representations as before. In general, replace all occurences of $\Phi_{pq}(g)$ above with $\Phi_{pq}(g) + N (1+x+\cdots + x^{pq})$ where $N$ is sufficiently large and divisible by $|G|$. $\endgroup$ – David E Speyer Dec 13 '18 at 1:16
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    $\begingroup$ Regarding the ettiquette question, it is usually better to start a new question which people have put significant effort into the old one. But, in this case, the new condition doesn't help anyway. $\endgroup$ – David E Speyer Dec 13 '18 at 1:29
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David Speyer's example shows that the answer is "no", in general, even in the cyclic case. However, here are a few general remarks about the case of finite Abelian groups.

Each linear character (ie, group homomorphism $G \to \mathbb{C}^{\times}$), say $\lambda,$ extends by linearity to an algebra homomorphism from $\mathbb{C}G \to \mathbb{C}$ (which we call $\lambda^{+}$ here).

When $G$ is Abelian, an element $u \in \mathbb{C}G$ is a unit of $\mathbb{C}G$ if and only $\lambda^{+}(u) \neq 0$ for each linear character $\lambda$ of $G.$

In a positive direction, the answer to your question is positive when $G$ is a finite Abelian $p$-group. If $u = \sum_{i}a_{i}g_{i}$ with $p$ not dividing $\sum_{i}a_{i}$ (each $a_{i} \in \mathbb{Z}$), then for each linear character $\lambda$ of $G$, we have $\lambda^{+}(u) \in \mathbb{Z}[\eta]$, but $\lambda^{+}(u) \equiv \sum_{i} a_{i} \not \equiv 0$ (mod $\pi$), where $\pi$ is the unique prime ideal of $\mathbb{Z}[\eta]$ containing $p$ ( $\eta$ being a primitive $p^{e}$-th root of unity where $G$ has exponent $p^{e}$). Then certainly $\lambda^{+}(u) \neq 0.$

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