Z8 is not an algebra group

Statement

The group cyclic group:Z8, defined as the cyclic group of order $2^3=8$, is not an algebra group.

Related facts

• Z4 is an algebra group
• Cyclic group of prime-square order is not an algebra group for odd prime

Proof

Note that, for a 2-group, being an algebra group is equivalent to being an algebra group over ${\mathbb{F}}_2$. So, we will prove that the group is an algebra group over ${\mathbb{F}}_p$.

Suppose $F={\mathbb{F}}_2$ and $N$ is a nilpotent associative algebra over $F$ such that the algebra group of $N$, which we denote $G$, is isomorphic to $\mathbb{Z}/8\mathbb{Z}$. $N$ is three--dimensional over $F$, so the unitization of $N$, i.e., the algebra $N+F$, is four-dimensional. The action of $G$ by left multiplication makes $G$ a subgroup of $GL(4,F)=GL(4,2)$ and hence of the Sylow subgroup UT(4,2) of upper triangular unipotent matrices. However, $UT(4,2)$ has exponent 4, so $G$, which has exponent 8, cannot be isomorphic to a subgroup of it.

References

• MathOverflow question: p-groups realisable as 1+J,where J is a nilpotent finite F-Algebra

Jack Schmidt's answer sketches the above proof.