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Let ${\rm SL}_2(\mathbb{Z}/8\mathbb{Z})$ be the special linear group over the finite ring $\mathbb{Z}/8\mathbb{Z}$. By How to calculate |SL2(Z/NZ)| , we know that the order of the group ${\rm SL}_2(\mathbb{Z}/8\mathbb{Z})$ is $384=2^7 \cdot 3$. By the Sylow theorem, there exists a matrix $A\in {\rm SL}_2(\mathbb{Z}/8\mathbb{Z}) $ of order $3$.

Is it true that the group ${\rm SL}_2(\mathbb{Z}/8\mathbb{Z})$ is generated by the set $\{B^{-1}AB~|~B\in {\rm SL}_2(\mathbb{Z}/8\mathbb{Z})\} $?

stupid boy
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1 Answers1

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I think the question is interesting, although the author doesn't respond to the comment about his own trying.

Let $G=SL(2,\mathbb{Z}_8)\}$.

  1. First note that $G'\neq G$, $G'$ is the commutator subgroup of $G$. Let $Q$ be a Sylow $2$-subgroup of group $G$. Since $|G:Q|=3$, there exists a homomorphism $G\to S_3$. Let $H$ be the kernel of this homomorphism.

    If $|G:H|=2$ or $|G:H|=3$, then $G'\leq H$
    If $|G:H|=6$, then $G'\leq F$, where $F$ is the preimage of a subgroup of order $3$ of $S_3$.

  2. Since all Sylow $2$-subgroups are conjugate in $G$, it suffices to find one element of order $3$ of $G'$. Then all elements of order $3$ lie in $G'$ and hence the set $\{B^{-1}AB\mid B\in G\}$, where $A$ is a element of order $3$, does not generate the group $G$.

  3. Let $$ A=\begin{pmatrix} -1 & -1\\1& 0\end{pmatrix},\ B=\begin{pmatrix} 3 & 2\\-1& -3\end{pmatrix}. $$ We have $$ B^{-1}=\begin{pmatrix} -3 & 6\\1& 3\end{pmatrix} $$ and $B^{-1}AB=A^2$ or $A^{-1}B^{-1}AB=A\in G'$.

That's it.

kabenyuk
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