Consider the invertible matrices in $\mathbb F_2^{n\times n}$ which are a multiplicative group structure. Is there a finite set of $2k$ (at a $k\in\mathbb Z_{\geq1}$ independent of $n$) generators for the group on the condition if $a$ is a generator $a\neq a^{-1}$ and $a^{-1}$ is a member of the set and is considered a different element compared to element $a$ (essentially the smallest set of no elements of order $2$ and is there an explicit example)?
$\begingroup$
$\endgroup$
11
-
5$\begingroup$ Silly answer: Since it's a finite group, just take all elements. Overkill answer: it is generated by two elements, compare mathoverflow.net/questions/59213/…. Systematic answer: the Gauss algorithm writes any matrix as a product of elementary matrices, so these form a set of generators of size $n^2-n$. $\endgroup$– Bertram ArnoldCommented Mar 10, 2021 at 7:54
-
2$\begingroup$ At the moment the question is trivial. Please edit it and ask a sensible question. $\endgroup$– Derek HoltCommented Mar 10, 2021 at 8:36
-
2$\begingroup$ There are certainly inverse closed generating sets of size $4$ that contain no involutions. This is true for all nonabelian simple groups. $\endgroup$– Derek HoltCommented Mar 10, 2021 at 8:46
-
2$\begingroup$ @1.. So is the question: "What is the smallest size of a generating set of $SL_n(F_2)$ containing no groupp elements of order 2? Is it independent of $n$?" If so, that's a reasonably sensible question, and I suggest you edit your question to make it that. $\endgroup$– HJRWCommented Mar 10, 2021 at 8:46
-
1$\begingroup$ What’s the point of the condition on inverses? Adding inverses to the generating set is completely useless: since this is a finite group, $a^{-1}$ is a power of $a$. $\endgroup$– Emil JeřábekCommented Mar 10, 2021 at 8:46
|
Show 6 more comments
1 Answer
$\begingroup$
$\endgroup$
1
OK, I'll answer the intended question! ${\rm SL}(2,2) \cong S_3$ and the answer to the question is no in that case, so assume that $n>2$.
Don Taylor wrote down explicit sets of two generators for the finite classical groups. Search for "D.E. Taylor Pairs of Generators for Matrix Groups". (These are the generators used by the software packages GAP and Magma.)
The generators for ${\rm SL}(n,2)$ are $x = I+E_{21}$ (a transvection with $1$ in the $(2,1)$ position), and a permutation matrix $y$ for the cycle $(1,2,3,\ldots,n)$.
Now $x$ has order $2$, but $xy$ and $y$ do not (proof left as exercise), so for our inverse-closed generating set we can take $\{y,y^{-1},xy,(xy)^{-1}\}$.
answered Mar 10, 2021 at 8:56
-
$\begingroup$ Is $SL(n, 2)$ identical to $SL(n,\mathbb F_2)$? Is transvection a matrix with single $1$ at said position? $\endgroup$– TurboCommented Mar 20, 2021 at 6:53