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I need to prove that $P=\left\{A\in M_2(\mathbb{R})\mid A^TXA = X\right\}$ is a group under matrix multiplication.

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    I guess it is worth emphasizing that in this case you didn't limit $A$ to a **group** of matrices, so you do need to check the existence of an inverse as in matt's answer. Sorry, I gave my knee-jerk response without checking all the details first.2011-12-30

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Closure: Let $A,B\in P$ then:

$(AB)^T X(AB) = B^TA^TXAB = B^T(A^TXA)B = B^TXB=X$

Therefore we have $(AB)\in P$.

Inverse: First observe that $X$ is invertible since $\det(X)=3\times 1-1\times 1 =2\not=0$. Now suppose $A\in P$ then we have:

$\det(A^TXA)=\det(X) \implies \det(A^T)\det(X)\det(A)=\det(X)\implies \det(A^T)\det(A)=1$

Since $\det(A^T)=\det(A)$ we have that $\det(A)\not=0$, that is, $A$ is invertible ($A^{-1}$ exists).

Moreover, $A^{-1}\in P$ since: $A^TXA=X\implies X=(A^T)^{-1}XA^{-1}\implies X=(A^{-1})^TXA^{-1}$

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    Good point showing that A^-1 does exist in the first place for a given A. That's not automatically true.2011-12-29
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For closure: $(AB)^T X AB = B^T A^T X AB = B^T X B = X$.

For inverses note that $\det(X) \neq 0$ and hence $\det(A^T X A) = \det(A^T) \det (X) \det(A) \neq 0$ and hence $A$ is invertible and so $(A^{-1})^T = (A^T)^{-1}$.

Use this to show that $A^{-1} \in P$.

Hope this helps.

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    @John Yes, it's what you have to show. The second line above shows exactly that. : )2011-12-30
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Hint: For closure, I'd suggest not to work out things explicitly, but use the fact that $(AB)^T=B^TA^T$. For the inverse, you can use the same property.

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    That just shows that the identity is contained in your set. So yes, that's OK.2011-12-29