# Let $a$ be the reflection of the plane $\mathbb{R}^2$ over the bisector of the odd quadrants [closed]

Let $$a$$ be the reflection of the plane $$\mathbb{R}^2$$ over the bisector of the odd quadrants (line with equation $$y = x$$), and let $$b$$ be the reflection of the same plane over the bisector of the even quadrants (line with equation $$y = -x$$).

a) Write the matrices of all elements of the group $$G = \langle a, b \rangle$$ (with the operation of composition of mappings) in the standard basis of the vector space $$\mathbb{R}^2$$: $$\mathcal{B} = \left\{ \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right\}.$$

b) Show that the given representation of the group $$G$$ is reducible: write the bases of the invariant subspaces $$U, V \subseteq \mathbb{R}^2$$, for which $$U \oplus V = \mathbb{R}^2$$ holds.

Attempt: (a)

The reflection over the line $$y = x$$ swaps the coordinates of any vector. Therefore, the matrix for $$a$$ is: $$a = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$

The reflection over the line $$y = -x$$ swaps the coordinates and changes their signs. Therefore, the matrix for $$b$$ is: $$b = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$$

What can I do now?

(b) Any idea on how to even start this task?

• For your standard basis, do you intend for them to be $1 \times 2$ row vectors (so they're dual covectors and matrices act on the right or equivalently by transpose) or is this just an issue with LaTeX notation? Commented Jul 12 at 19:56
– Shaun
Commented Jul 12 at 19:57

For part (a), note that $$G = \langle a, b \rangle$$ indicates the subgroup generated by $$a$$ and $$b$$ (in which larger group?). So, in general, this can be rather large, since any word we write down in the alphabet $$\{a, b, a^{-1}, b^{-1}\}$$ is an element of $$G$$, e.g., $$1, a, b, a^{-1}, b^{-1}, a^2, ab, ba, b^2, a^3, a^2b, aba, ab^2, ba^2, bab, b^2a, b^3, a^4, \ldots$$ If all of these are distinct, then we have a free group, but that is not the case here. In fact, this group is much smaller!

To start with, we can calculate $$c = ab$$ and verify that

• each of $$a, b, c$$ has order $$2$$, i.e., $$a^2 = b^2 = c^2 = 1$$
• they all commute, i.e., $$ab = ba$$, etc.

Can you see how this means that any word in $$G$$ is actually just one of $$\{1, a, b, c\}$$? Can you write down the $$4 \times 4$$ multiplication table for $$G$$? Do you recognize this structure from elsewhere? (It's isomorphic to a well-known group $$V$$, and you'll often hear this stated as "$$G$$ is the group $$V$$" by implicitly invoking the isomorphism.

• Can you just tell me, what is the group $V$ here, since I cannot figure it out?
• Did you verify that $a^2 = 1$ and $b^2 = 1$? Just multiply the matrices. Once you have that, any word in $a$s and $b$s cannot have an exponent greater than $1$. So the only distinct nontrivial elements are $a, ab, aba, abab, ababa, \ldots$ and $b, ba, bab, baba, babab, \ldots$. Then show that $ab = ba$. So now, order doesn't matter, and all we have left are $1$, $a$, $b$, and $ab$ (which is of course $ba$). Call that last one $c$. Now, it's easy to see how they multiply. Commented Jul 15 at 20:46