# If $K[Q_8]\cong K[D_8]$, char $K=p$ odd, $p=?$

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 $p$.