$3\times3$ invertible matrix in $\mathbb{F}_2$ such that $M^7 = I$

Question

Show that there exists a $3\times 3$ invertible matrix $M\neq I_3$ with entries in the field $\mathbb{F}_2$ such that $M^7 =I_3$.

Attempt: $M^7 =I$ can be factorized as $(M- I) ( M^6 + M^5 +...+M+I)=0$ but I don't think it will prove the existence of required M. Trying to find an M by hit and trial method seems a bad idea.

Can you please suggest some elegant way of proving it?

$\text{GL}_3(\mathbb{F}_2)$ has order $168$ (see https://math.stackexchange.com/questions/34271/order-of-general-and-special-linear-groups-over-finite-fields). Then use Cauchy's theorem since $168$ is a multiple of $7$. — lhf, Sep 07 '21 at 11:00

score 4 · Answer 1 · answered Sep 07 '21 at 11:10

4

Since $x^7-1=x^7+1=(x+1)(x^3+x+1)(x^3+x^2+1)$, it suffices to take $M$ to be a companion matrix of $x^3+x+1$ or $x^3+x^2+1$.

answered Sep 07 '21 at 11:10

user1551

139,064

Why should I look at companion matrix? I mean what is the use of concept of companion matrix here. – Sep 11 '21 at 07:24
@No-One Because for every monic degree-$n$ polynomial $p$, there exists a companion matrix whose characteristic polynomial is precisely $p$. – user1551 Sep 11 '21 at 10:41

$3\times3$ invertible matrix in $\mathbb{F}_2$ such that $M^7 = I$

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