Let $G$ be a group of order $\le 4$. Find the unit group of the group ring $\mathbb Z[G]$.

Trivially, if $|G|=1$, then the unit group is $\{\pm e\}$, where $e$ is the identity element fo $G$.

For $|G| \ge 2$, what should I do? If $|G|=2$, let $a$ be a nonidentity element of $G$, and I let $(ma+ne)(pa+qe)=e$, and tried a long, messy calculation and found the answer.

However, I totally stuck in the case $|G| \ge 3$. How should I find all the units in this case? I already know that a group of order 3 is cyclic, and a group of order 4 is either cyclic or isomorphic to Klein-4 group. Also, for the case $|G|=2$, is there more elegant solution without the brutal computation(just what I did)?

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    $\begingroup$ As I noted in my (now deleted) answer, there is an isomorphism of $\mathbb{Z}$-algebras $\mathbb{Z}[G] \to \mathbb{Z}[X]/\langle X^{n}-1\rangle$. As Dustan Levenstein carefully pointed out, one cannot apply the Chinese Remainder Theorem here, so this isomorphism is not as helpful as I imagined. However, you may see how to use this, so I thought I'd mention it nevertheless. $\endgroup$ – Alex Wertheim Feb 9 '17 at 2:05
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    $\begingroup$ Actually, I don't think Alex's idea should be too difficult to rescue. If you find an explicit isomorphism of $\mathbb C[G]$ with a product of matrix algebras, then it's easy to identify the units in that picture, and then intersect that group of units with $\mathbb Z[G]$. The difficulty lies in finding the isomorphism and pulling the group of units back through the isomorphism. $\endgroup$ – Dustan Levenstein Feb 9 '17 at 2:17
  • $\begingroup$ That said, if this is a homework problem, there's a decent chance that that's not the intended solution. $\endgroup$ – Dustan Levenstein Feb 9 '17 at 2:19
  • $\begingroup$ This is not a homework problem, but a problem from graduate entrance exam. $\endgroup$ – bellcircle Feb 9 '17 at 2:20
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    $\begingroup$ In that case, this approach is probably fair game. $\endgroup$ – Dustan Levenstein Feb 9 '17 at 2:22

This answer is the approach outlined in the comments, with the caveat that passing all the way to $\mathbb C$ is overkill. There is one more piece of difficulty which I failed to notice in the comments: a unit in $\mathbb C[G]$ which is an element of $\mathbb Z[G]$ is not necessarily a unit in $\mathbb Z[G]$.

We use the fact that $\mathbb Q[X]/(X^n-1)$ breaks down into a product of Cyclotomic extensions of $\mathbb Q$, by the Chinese Remainder Theorem. I'll write $C_n$ for the cyclic group of order $n$, to emphasize that these groups are being written multiplicatively.

  • $G = C_2$. We have $$\mathbb Q[G] = \mathbb Q[X]/(X^2-1) = \mathbb Q[X]/(X-1) \times \mathbb Q[X]/(X+1).$$ Writing $C_2 = \langle \sigma\rangle$, We have $\sigma \mapsto (1, -1)$ under this isomorphism (think of reducing "modulo $\sigma=1$" in the left term, and modulo $\sigma=-1$ in the right), and $$a+b\sigma \mapsto (a, a) + (b, -b) = (a+b, a-b).$$ In particular, $\mathbb Z[G]$ is a (strict) subring of $\mathbb Z \times \mathbb Z$ under this presentation, so to be a unit, it is necessary that $$a+b, a-b \in \{\pm 1\}.$$ It follows that $a, b \in \{1, 0, -1\}$, which doesn't leave many cases to check.
  • $G = C_3$. We have $$\mathbb Q[G] = \mathbb Q[X]/(X^3-1) = \mathbb Q[X]/(X-1) \times \mathbb Q[X]/(X^2+X+1).$$ Write $\zeta$ for the image of $X$ in $\mathbb Q[X]/(X^2+X+1)$. Letting $G = \langle \sigma\rangle$, we have $\sigma \mapsto (1, \zeta),$ and $$a+b\sigma + c\sigma^{-1} \mapsto (a+b+c, a+b\zeta+c\zeta^{-1}).$$ This embeds $\mathbb Z[G]$ into $\mathbb Z[X]/(X-1) \times \mathbb Z[\zeta]$. The units in $\mathbb Z[\zeta]$ can be determined using the Galois-theoretic norm, which happens to be the same as the complex norm in this case: $$(a+b\zeta+c\zeta^{-1})(a+b\zeta^{-1}+c\zeta) = a^2+b^2+c^2+(ab+ac+bc)(\zeta+\zeta^{-1})$$ $$= a^2+b^2+c^2-(ab+ac+bc).$$ So it is necessary that $$a+b+c, a^2+b^2+c^2-(ab+ac+bc) \in \{\pm 1\}.$$ To solve this system of equations, it is helpful to notice that $$(a+b+c)^2 = a^2+b^2+c^2+2(ab+ac+bc)$$ must equal $1$, and the difference between that and $a^2+b^2+c^2-(ab+ac+bc)$ is $3(ab+ac+bc)$, which is $3$ times an integer. Since the difference between $1$ and $-1$ is only $2$, it follows that $$ab+ac+bc=0$$ and $$a^2+b^2+c^2 = \pm 1,$$ which is enough to determine what the units are.

I'll leave the remaining cases $G = C_4$ and $G= C_2 \times C_2$ to you to do; I believe the former should be easier than $C_3$ was, since it is easier to compute norms in $\mathbb Z[i]$. For $C_2 \times C_2$, note that $$R[C_2 \times C_2] = R[X, Y]/(X^2-1, Y^2-1),$$ where $R$ is any commutative ring.

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  • $\begingroup$ Excellent! Nicely done, +1. $\endgroup$ – Alex Wertheim Feb 9 '17 at 7:11

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