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If i have a specific matrix group with matrices $A,B,C,D$ where $Α$ is the identity element and also $BD=A$,$DB=A,C^2=A,A^2=A$. That means that $C$ and $A$ are inverse elements of their selfs.

Does this creates problems to it's requirements for being a group?

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    That's fine. Your matrix group is cyclic of order 4.There even exist groups of very large or even infinite orders where every element is self-inverse. – 2017-02-19

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No, there's no problem. It's perfectly possible to have elements of a group that are their own inverses. For instance, the complex numbers $A=1$, $B=i$, $C=-1$, and $D=-i$ satisfy all the equations you've written and form a group (even a group of matrices, if you consider them as $1\times 1$ matrices).

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    Or even $2\times 2$ matrices if you consider the usual matrix representation. – 2017-02-19
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They can be:

$$ A=\begin{bmatrix} 1&0\\ 0&1 \end{bmatrix} \qquad C=\begin{bmatrix} -1&0\\ 0&-1 \end{bmatrix} \qquad B=\begin{bmatrix} 0&-1\\ 1&0 \end{bmatrix} \qquad D=\begin{bmatrix} 0&1\\ -1&0 \end{bmatrix} $$