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Denote $Q_8$ to be the quaternion group, and $D_8$ to be the dihedral group with order 8, then we know that the group algebra $\mathbb{C}[Q_8]\cong \mathbb{C}[D_8]$ since $Q_8$ and $D_8$ have the same character table.

Now, by Maschke's theorem, if $K$ is a finite field with characteristic odd prime $p$, the group algebras $K[Q_8], K[D_8]$ are also semisimple algebras,

Can we also have that $K[Q_8]\cong K[D_8]$? If it maybe, then $p=?$


I only know that by Wedderburn' theorem, the algebra (as a ring) decomposes as the product of fininte matrix rings over division rings, but how to proceed? Any hint?

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This will be true as long as $K$ is a splitting field for $G$, i.e. as long as all irreducible representations of $G$ over $\bar{K}$ can be defined over $K$. That's because these representations will be more or less "the same" as the complex ones. Concretely, every complex representation of $Q_8$ can be realised over $\mathbb{Z}[i]$. Pick a prime ideal $\mathfrak{p}$ in $\mathbb{Z}[i]$ and reduce all the matrix entries modulo $\mathfrak{p}$. You will get an irreducible representation over the residue field $K$ of $\mathfrak{p}$. If the prime ideal $\mathfrak{p}$ is split in $\mathbb{Z}[i]$, then the characteristic of $K$ is $p\equiv 1\pmod{4}$ and $K=\mathbb{F}_p$. If the prime ideal is inert, then the characteristic of $K$ is $p\equiv 3\pmod{4}$ and $K=\mathbb{F}_{p^2}$.

Intuitively, you can think of it like this: if $p\equiv 1\pmod 4$, then $\mathbb{F}_p$ contains the fourth roots of unity, so there is nothing stopping you from realising the usual complex representations over $\mathbb{F}_p$, by simply replacing $\pm i$ with these fourth roots of unity. If $p\equiv 3\pmod 4$, then to realise your representations, you need to pass to the quadratic extension to acquire the fourth roots of unity.

In summary, for any odd $p$, there is a finite field $K$ of characteristic $p$ such that $K[Q_8]\cong K[D_8]$. But depending on $p$, this $K$ might have to be of order $p^2$, rather than $p$.

One can also express all of this in terms of Wedderburn and division rings. The upshot is that as soon as you pass to the splitting field, your division rings are all equal to this base field.

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  • $\begingroup$ Thank you very much! It's very impressing! $\endgroup$ – ougao Jan 7 '13 at 18:49
  • $\begingroup$ @ougao You are welcome! $\endgroup$ – Alex B. Jan 8 '13 at 9:37

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