Classifying a set closed under two unary operators

Suppose we have a single square $S$ with distinct colors at its four vertices. We define two operators: the rotation operator $R$, which rotates the square $90^\circ$ clockwise, and the transpose operator $T$, which flips the square along its main diagonal (from the top left to bottom right vertex).

We apply $R$ and $T$ to the square repeatedly in any order we like and for as many times we like. We get a collection of squares, such that when either $R$ or $T$ is applied to any one element, we get back an element in the same collection.

The closest thing that comes into my mind as to what the above collection should be called, is "a group closed under $R$ and $T$ with generator $S$" - however the definition of a group only allows for one group law, and it also requires the group law to be a binary operator.

How should such a collection be classified?

• In fact there is a group, but it is generated by $R$ and $T$. – ajotatxe Apr 4 '15 at 15:21

$R$ and $T$ are actually elements of a group and the group they generate is the group of symmetries of the square. There are 8 elements of this group including the identity which leaves the entire square fixed. The binary operation here is function composition, which is to say, $R*T$ is the motion that results in applying $T$ to the square and then applying $R$ to the square.