Given the Quaternion group $$Q_8=\{\pm1,\pm i,\pm j,\pm k\}$$ with the product: $1$ is the unit, $-1$ commutes with all elements, $(-1)^2=1$ and $i^2=j^2=k^2=ijk=-1.$ The group ring $\mathbb{R}[Q_8]$ has dimension $8$ (as an algebra over $\mathbb{R}$). The elements of $Q_8$ all have multiplicative inverses. Why is this not a division algebra over $\mathbb{R}$?

The easy solution is: Such an division algebra has dimension $1$, $2$ or $4$ (theorem of Frobenius).

Can this be shown without using Frobeniu's theorem about division algebras ober the real numbers?

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    $\begingroup$ The group ring $\Bbb R Q_8$ is a little funky because, e.g. $-2(i)\neq 2(-i)$. I suspect this quirkiness might be useful in getting a contradiction. $\endgroup$ – Eric Stucky Oct 23 '16 at 17:25
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    $\begingroup$ I would recommend rephrasing the question, because it is confusing to have a group element labelled $-1$ that is not the additive inverse of $+1$ in $\Bbb{R}[\mathrm{Q}_8]$. Perhaps try something like $\mathrm{Q}_8=\{e,\overline{e},i,\overline{i},j,\overline{j},k,\overline{k}\}$, so that each of the eight can be used as a basis element of the group ring. We have that $-e\ne\overline{e}$. To show what you want, you only have to show the existence of a nontrivial zero divisor. $\endgroup$ – qman Oct 23 '16 at 17:40
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    $\begingroup$ Why would associativity fail if it is a ring $\endgroup$ – user369147 Oct 23 '16 at 18:07
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    $\begingroup$ Associativity wouldn't fail. I was going to query the same. $\endgroup$ – qman Oct 23 '16 at 18:09
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    $\begingroup$ See math.stackexchange.com/questions/1978912/… for a nice and general way to see this. $\endgroup$ – Tobias Kildetoft Oct 23 '16 at 18:13

There are extremely simple ways to see this.

You could observe that group rings are almost never simple rings.

Another good thing for your toolbox for finite groups $G$ is that if $G$ has a nontrivial normal subgroup, $\mathbb R[G]$ isn't even a domain. The quaternion group has, to put it mildly, lots of normal subgroups.

To see this, note that the sum of elements in a finite nontrivial normal subgroup creates a nontrivial central idempotent that splits the ring in two pieces. If the normal subgroup $H$ had order $k$, then $\frac1k\sum_{h\in H}h$ is a central idempotent. Consequently $eR\oplus (1-e)R$ splits the ring into the product of two rings. Or more simply, $e(1-e)=0$ proves there are zero divisors.

This only relies on $|H|$ being a unit in the coefficient ring. There are still some related tricks to this in rings of positive characteristic, but it's more than sufficient already for this problem. If the order of the subgroup is a divisor of the characteristic of the ring, then the sum of the elements in the subgroup is nilpotent, and it is not a domain (or division ring) in that case either.

  • $\begingroup$ Let us look at your first point. So the idea is to see that there is a two-sided ideal in $\mathbb{R}[Q_8]$, which is not zero or equal to $\mathbb{R}[Q_8]$. Is we have this, $\mathbb{R}[Q_8]$ can't be a field since a field has only the two trivial ideals. So there are two questions left: #1 What is a possible ideal and #2 Why it is not a division algebra if it it not a field? $\endgroup$ – user369147 Oct 23 '16 at 18:58
  • $\begingroup$ @user114179 a ring is called simple if it only has trivial ideals. Fields $\subseteq$ division rings $\subseteq$ simple rings. Therefore if it's not simple, it can't be a division ring either . $\endgroup$ – rschwieb Oct 23 '16 at 19:15
  • $\begingroup$ Yeah, but we are talking about division $\textit{algebras}$. But if it is not a division ring, it can't be a division algebra, right? $\endgroup$ – user369147 Oct 23 '16 at 19:17
  • $\begingroup$ @user114179 By the definition in thinking of, a division algebra is a priori a division ring. Does your definition of division algebra not assume associativity? $\endgroup$ – rschwieb Oct 23 '16 at 21:11
  • $\begingroup$ It does. A ring is associative by definition. But what is the not trivial ideal we need? $\endgroup$ – user369147 Oct 23 '16 at 21:24

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