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Let $H$ be the special linear group. $H=\{M \in M_n(\mathbb{K}): det(M) = 1\}$. Is $H$ a subgroup of the orthogonal group?

Counterexample:

Let $A = \left( \begin{array}{ccc} 1 & 1 \\ 0 & 1 \\ \end{array} \right)$

Then $det(A) = 1$. But $AA^T \neq Id$. So $A$ isn't orthonormal.

Is that a valid counterexample?

karnan
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    The statement is ambiguous, since there is no such thing as the orthogonal group. But, yes, that proves that $SL(n,\mathbb K)\varsubsetneq O(n,\mathbb K)$. – José Carlos Santos Apr 16 '20 at 11:41
  • I wasn't sure about this either. I assumed that they mean all matrices which are orthogonal. Thanks! – karnan Apr 16 '20 at 11:47

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