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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)?

Turbo
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    Silly answer: Since it's a finite group, just take all elements. Overkill answer: it is generated by two elements, compare https://mathoverflow.net/questions/59213/generating-finite-simple-groups-with-2-elements. Systematic answer: the Gauss algorithm writes any matrix as a product of elementary matrices, so these form a set of generators of size $n^2-n$. – Bertram Arnold Mar 10 '21 at 07:54
  • @BertramArnold Can you explicitly provide the two generators or any set of $2k$ generators? – Turbo Mar 10 '21 at 07:59
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    At the moment the question is trivial. Please edit it and ask a sensible question. – Derek Holt Mar 10 '21 at 08:36
  • The request seems to be for a generating set that contains no elements of order $2$. OK, take all elements in the group of order. – Derek Holt Mar 10 '21 at 08:42
  • Smallest generating set of no elements of order $2$. – Turbo Mar 10 '21 at 08:44
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    There are certainly inverse closed generating sets of size $4$ that contain no involutions. This is true for all nonabelian simple groups. – Derek Holt Mar 10 '21 at 08:46
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    @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. – HJRW Mar 10 '21 at 08:46
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    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 $a$. – Emil Jeřábek Mar 10 '21 at 08:46
  • ... and then @DerekHolt can write a post explaining why the answer is 4! :) – HJRW Mar 10 '21 at 08:48
  • @emiljerabek Order should not be $2$. – Turbo Mar 10 '21 at 08:49
  • @HJRW Correct version. – Turbo Mar 10 '21 at 08:49

1 Answers1

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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 $\{y,y^{-1},xy,(xy)^{-1}\}$.

Derek Holt
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  • Is $SL(n, 2)$ identical to $SL(n,\mathbb F_2)$? Is transvection a matrix with single $1$ at said position? – Turbo Mar 20 '21 at 06:53