Prove that $(G, \circ)$ is a group

Question

Given that $F_1, F_2, F_3, F_4$ are applications of $F_i: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ which are defined as $F_1: (x, y) = (x, y), F_2: (x, y) = (-x, y), F_3: (x, y) = (x, -y), F_4: (x, y) = (-x, -y)$. If $G = \{F_1, F_2, F_3, F_4\}$, demonstrate that $(G, \circ)$ is a group.

I know that I have to prove the internal law, associative rule, the existence of the neutral element and the existence of the symmetric elements for all $F_i \in G$ but I am not sure where to begin as I have not worked with sets of functions and the problem does not state the definition of $\circ$.

Is there a predefined representation of $\circ$? Also, how do I go about solving this problem?

Answer 1

Usually, it is always good in such situations to find a bijective map (between sets) from a "known group" to the present structure, and show that the operation of the "known group" is transported into the new (to-be group) structure. Then the new structure is a group by transport of structure.

In the given case, the given applications correspond to the left multiplication with the invertible matrices of type $2\times 2$ over $\Bbb R$, $$ \begin{bmatrix}1&0\\0&1\end{bmatrix}\ ,\ \begin{bmatrix}-1&0\\0&1\end{bmatrix}\ ,\ \begin{bmatrix}1&0\\0&-1\end{bmatrix}\ ,\ \begin{bmatrix}-1&0\\0&-1\end{bmatrix}\ . $$ The set of all invertible matrices ype $2\times 2$ over $\Bbb R$ for a group, the general linear group.

The above four matrices form a stable subset, closed w.r.t. multiplication and taking inverse, so it forms a subgroup.

The given structure is by transport of structure also a subgroup.

Note: It is isomorphic to $(\Bbb Z/2)\times (\Bbb Z/2)$.

Note: This solution is not always welcome, the problem setter really wants often a bloody check of all axioms... It happened to me. (In this case, the adapted objection would be as follows. The problem asked to show it is a group. The above shows it is a subgroup of some ?group?. Then the examiner really wanted to show that the bigger ?group? is a group. Olympiads work like this.)

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Later note: Let us make things even more explicit. We have a set $\{F_1,F_2,F_3,F_4\}$ with an operation, $\circ$, composition in our case.

We have a set of four matrices $\{A_1,A_2,A_3,A_4\}$, all of them diagonal with entries $\pm1$ with a stable operation, the matrix multiplication. (The multiplication of two such matrices with diagonal entries $\pm 1$, respectively the inverse... is again a matrix with entries $\pm 1$, stability w.r.t. multiplication and taking inverse.)

The group of all invertible matrices is indeed a group. (I took an other structure, not the bijective functions with composition to have a less tautological situation.)

The set $\{A_1,A_2,A_3,A_4\}$, with inherited operations is a subgroup.

We have a bijection $F_k\leftrightarrow A_k$ of sets, and it is compatible with the operations, $F_k\circ F_l\leftrightarrow A_k\cdot A_l$ in case we have correspondences $F_k\leftrightarrow A_k$ and $F_l\leftrightarrow A_l$. (To avoid showing this, use functions instead of matrices, that tautological situation.)

So the question deals with a "transported group structure", which is a group.

Answer 2

Since $\operatorname {Sym}(X)$ is a group for every set $X$, and all four $F_i$'s are bijections on $X=\mathbb R^2$, it suffices to prove that $G$ (which is then a nonempty finite subset of a group) is closed under map composition. And in fact (let me omit all the "$\circ$"): \begin{alignat}{1} &F_1F_k=F_k, k=1,2,3,4 \\ &F_kF_1=F_k, k=1,2,3,4 \\ &F_k^2=F_1, k=1,2,3,4 \\ &F_2F_3=F_4 \\ &F_2F_4=F_3 \\ &F_3F_4=F_2 \\ &F_3F_2=F_4 \\ &F_4F_2=F_3 \\ &F_4F_3=F_2 \\ \end{alignat} Btw, since all the nontrivial elements of $G$ have order $2$, $G\cong C_2\times C_2$.