Let $F$ be any finite field of characteristic different from $2$. I have to prove $F(Z_2\times Z_2)\cong F\oplus F\oplus F \oplus F.$ I know that $F(Z_2\times Z_2)\cong F(Z_2)\otimes_F F(Z_2)$ and $F(Z_2)\cong F+F, $ here $\otimes$ is the tensor product and $F(G)$ denotes the group algebra of the group $G$ over the field $F$. How to proceed further? Please help. Thanks.


Let's write the Klein 4-group multiplicatively as $\mathbb Z_2\times\mathbb Z_2$ as $\{1,a,b,ab\}$ with the usual rules regarding their multiplication (that is, they all square to $1$).

A useful fact (and easy exercise) is that if $H$ is a finite normal subgroup of a group $G$, and the order of $H$ is a unit in $F$, then $\frac{1}{|H|}\sum_{h\in H}h$ is a central idempotent of $F(G)$.

Let $e=\frac{1+a}{2}$, $g=\frac{1+b}{2}$ and $f=\frac{1+a+b+ab}{4}$.

These are all central idempotents according to the previous proposition.

Now one can check that $\{f, e-f, g-f, 1-e-g+f\}=X$ is a complete set of orthogonal idempotents for $F(G)$ in the sense that $\sum_{x\in X}x=1$ and $xy=0$ for all $x,y\in X$.

In such a case, $F(G)\cong \prod_{x\in X}xF(g)x$ as rings via the obvious map.

Maschke's theorem says that the ring is semisimple, and the Wedderburn theorem says $F(G)$ is a finite product of semisimple rings, but considering that $F(G)$ is commutative and we already have decomposed it into $4$ rings, the only possibility is that each of the rings is $F$.

  • $\begingroup$ @rachwieb I am confused by the tensor product ... $\endgroup$ – neelkanth Jan 24 at 17:03
  • $\begingroup$ How to handle tensor product in between ... $\endgroup$ – neelkanth Jan 24 at 17:04
  • $\begingroup$ @neelkanth Yes, good question. I'm pretty sure we can get rid of it, I'm just trying to think of the easiest way to explain. Actually, I might fall back on applying my argument to the entire group, rather than focusing on the piece $F(\mathbb Z_2)$. $\endgroup$ – rschwieb Jan 24 at 17:06
  • $\begingroup$ ok thanks ...... $\endgroup$ – neelkanth Jan 24 at 17:07
  • $\begingroup$ @neelkanth I rewrote using a concrete argument. $\endgroup$ – rschwieb Jan 24 at 18:28

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