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A matrix satisfies

$$X(a)SX(a)^T = S$$

Where $X^T$ is the transpose of the $2 \times 2$ matrix $X$ and $S$ is a matrix $$\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$$ I have a group where I think all matrices satisfy this property, which is probably unusual and has a special name. What is so important or useful about this property? This sort of thing is exceptionally hard to google, so thanks for any help!

Edit: Original question is to show the matrices $X(a)$ form a group, and to calculate $$X(a)SX(a)^T$$ and state why the group is special.

X is the $2 \times 2 $ matrix $\left( \begin{array}{cc} cosh(\alpha) & sinh(\alpha)\\ sinh(\alpha) & cosh(\alpha) \end{array} \right)$ For $\alpha \in \mathbb{R}$.

Then $$X(\alpha)S = \left( \begin{array}{cc} cosh(a) & sinh(a)\\ sinh(a) & cosh(a) \end{array} \right)\left( \begin{array}{cc} 1 & 0\\ 0 & -1 \end{array} \right)$$

$$= \left( \begin{array}{cc} cosh(a) & -sinh(a)\\ sinh(a) & -cosh(a) \end{array} \right)$$

$$X(a)SX^T(a) = \left( \begin{array}{cc} cosh(a) & -sinh(a)\\ sinh(a) & -cosh(a) \end{array} \right) X(a)$$ since $X(a)= X^T(a)$

$$= \left( \begin{array}{cc} cosh^2(a) -sinh^2(a)& cosh(a)sinh(a)-sinh(a)cosh(a)\\ sinh(a)cosh(a)-cosh(a)sinh(a) & -cosh^2(a) + sinh^2(a)\end{array} \right)$$ Using $cosh^2 - sinh^2 = 1$, then this matrix is $S$.

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    If $X$ is invertible then you can say that $S$ is [congruent](https://en.wikipedia.org/wiki/Matrix_congruence) to $I$ (and vice-versa). – 2017-02-04
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    Well, since the right hand side of the equation is positive definite, and the left hand side is undefined, the equation can't hold for any matrix $X$ (with exception of the case when the main field characteristic is 2). – 2017-02-04
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    @SergeiGolovan Never come across the term main field characteristic, but I've only shown this to hold for the group of matrices $$X(\alpha) = \left( \begin{array}{cc} \cosh(\alpha) & \sinh(\alpha ) \\ \sinh(\alpha ) & \cosh(\alpha ) \\ \end{array} \right)$$ where $\alpha$ is real. – 2017-02-04
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    If $X=[1 \ 0; 0\ i]$ and you really use transpose, not hermetian conjugate, then you have an example, but the concept seems rather weird? No real matrix can verify the equation: The determinant on the LHS being negative in that case. – 2017-02-04
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    Maybe it will be helpful if you state the original question also along with your work. – 2017-02-04
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    In your example, you also get $S$ on the RHS, not I – 2017-02-04
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    @H.H.Rugh Having checked my calculation I still get $I$. – 2017-02-04
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    Please check your calculation once again. – 2017-02-04
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    Your lower right corner has the opposite sign (should be $\sinh^2-\cosh^2$) – 2017-02-04
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    @H.H.Rugh Ah, there it is. Thanks! – 2017-02-04

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It looks as if your group, with the hyperbolic cosh, sinh example verifies: $$ X S X^T = S$$ with $S=\left( \begin{matrix} 1 & 0 \\ 0& -1\end{matrix} \right)$. The matrix $S$ defines a metric with signature $(1,1)$ (one positive, one negative eigenvalue). Your group is an isometry of that metric.

${\Bbb R}^2$ equipped with this metric is called the Minkowski plane and is used e.g. in Einstein's special relativity. You may similarly look in ${\Bbb R}^4$ at the metric: ${\rm diag} (-1,1,1,1)$ and the isometry groups associated (which then becomes a mixture of hyperbolic cos/sin and standard cos/sin)

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    Ok I've added my workings because I can't see where I'm going wrong. It's probably something incredibly stupid and obvious, but I can't see it! Is my transpose right? That would be my first thought... – 2017-02-04
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    @Karacoreable , as a complement to Rugh's answer, you may consider that $S$ is one of the square roots of $I$. – 2017-02-04