2018
Том 70
№ 12

# Tsikunоv I. K.

Articles: 1
Article (Russian)

### Silov $p$-subgroups of orthogonal and simplex groups of a hyperbolic space

Ukr. Mat. Zh. - 1963. - 15, № 3. - pp. 290-298

The authors consider the construction of Silov $p$-subgroups of a group of isometrics (linear transformations of a space preserving the metric) of a bilinear-metric space with orthogonal or simplex metrics, consisting of an orthogonal sum of two-dimensional subspaces of the form $$V_{2n} = \langle N_1, M_1\rangle \perp \langle N_2, M_2\rangle \perp ...\perp \langle N_n, M_n\rangle ,$$ where $\langle N_i, M_i \rangle$ is a linear shell of the vectors $N_i$ and $M_i$, $N_i^2 = M_i^2 = 0,\; N_i \cdot M_i = 1$ in the sense defined in the $V_{2n}$ scalar product (metrics).

The results are formulated as the following theorems.
Theorem 1. In order that subgroup $S$ of a group of isometrics of space $V_{2n}$ should be a Silov $p$-subgroup, the following conditions are necessary and sufficient:
a) subgroup $S$ must have an invariant maximal isotropic subspace $U_n \subset V_{2n}$
b) subspace $S$ must induce in $U_n$ a Silov $p$-subgroup of its full linear group.

Theorem 2. The Silov $p$-subgroup $S$ of a group of isometrics of space $V_{2n}$ is a semidirect product of the Silov $p$-subgroup of the full linear group of the vector space $U_n$ and the normal divisor $H$, which in the orthogonal metrics is isomorphic to the additive group of obliquely symmetric matrices of order n, in simplex metrics — to the additive group of symmetrical matrices of order $n$. b) subspace $S$ must induce in $U_n$ a Silov $p$-subgroup of its full linear group.

Theorem 2. The Silov $p$-subgroup $S$ of a group of isometrics of space $V_{2n}$ is a semidirect product of the Silov $p$-subgroup of the full linear group of the vector space $U_n$ and the normal divisor $H$, which in the orthogonal metrics is isomorphic to the additive group of obliquely symmetric matrices of order n, in simplex metrics — to the additive group of symmetrical matrices of order $n$.