Timeline for Why are all involutions conjugate in the special linear group of degree 2?
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| May 15, 2014 at 7:20 | comment | added | Geoff Robinson | Yes, it is easy to see directly. Setting $q = 2^{n},$ the normalizer of a Sylow $2$-subgroup has order $q(q-1),$ and contains a cyclic subgroup of order $q-1$ which permutes the non-identity elements of the Sylow $2$-subgroup transitively under conjugation. | |
| May 15, 2014 at 1:57 | history | answered | user94741 | CC BY-SA 3.0 |