Show that $(K,\circ)$ is a group.

Let $K$ be the set of functions defined by : $f : \mathbb C \times \mathbb C \to \mathbb C \times \mathbb C$, such that $\exists a \in \mathbb C, \exists b \in \mathbb C$, $a,b$ not simultaneously equal to zero with : $f(u,v) = (au+bv,-\bar b u+\bar a v$)

Q:Show that $(K,\circ)$ is a group.

I was able to show $f_1 \circ f_2 \in K$ (Closure) and $(f_1 \circ f_2)\circ f_3 = f_1 \circ (f_2 \circ f_3)$ (Associativity) . but i'm not familiar nor able to show the symmetry nor the identity element of this set.

I'd appreciate any help i can get that would push me in the right direction! thanks.

• Why has this question been downvoted? The question is good and it has been stated clearly. Also, the OP has attempted the problem and successfully solved one part of the problem (closure) which happens to of moderate difficulty. – stressed out Dec 9 '17 at 13:29
• It's completely fine by my side! i got your help which actually awakened me from my dumb state (haha!), which helped me understand what i'm supposed to do with this kind of problems! thank you. – Manletta Dec 9 '17 at 14:07

According to you, you have already shown the closure of such functions. Also, associativity is always true for composition of functions, no matter what. To prove that $K$ is a group under composition of functions, we have to show the existence of an identity element in the group, and the existence of the inverse of any element in the group.

The identity function will be the identity element of $K$ and it is in $K$ because for $a=1,b=0$, we have $f(u,v)=(u,v)$. Therefore, the identity function is in the set $K$.

Now to find the inverse of $f$, you have to solve the following system of equations in $\mathbb{C}^2$, $$au+bv=c$$ $$-\bar{b}u+\bar{a}v=d$$

You want to solve it for the unknown $u$ and $v$. The determinant is $|a|^2+|b|^2$ which is not $0$ because $a,b$ are not simultaneously $0$. Therefore, the system has a unique solution. Let's find it:

$$\pmatrix{u \\ v} = \frac{1}{|a|^2+|b|^2}\pmatrix{\bar{a} && -b \\ \bar{b} && a}\pmatrix{c \\d}$$

$$\pmatrix{u \\ v} = \pmatrix{ \frac{\bar{a}c-bd}{|a|^2+|b|^2} \\ \frac{\bar{b}c+ad}{|a|^2+|b|^2}}$$

If you set $$r=\frac{\bar{a}}{|a|^2+|b|^2}$$ $$s=\frac{-b}{{|a|^2+|b|^2}}$$

You obtain, $$f^{-1}(c,d)=(rc+sd,-\bar{r}c+\bar{s}d)$$ which is of the given form and therefore, belongs to $K$.

• Thank you very much for your answer. If i'm getting this right, i'm supposed to show that $f^{-1} \circ f = f \circ f^{-1} = Id$ for symmetry right? – Manletta Dec 9 '17 at 12:30
• I don't know what you mean by symmetry. You're probably using a term that is standard in physics, but not standard in mathematics. Your question also seems to be of physical interest. If you show that $f^{-1}$ of the function is of the given form in $K$, then since it is the inverse of $f$, it is certainly true that composition of $f^{-1}$ with $f$, from left or right, should give the identity. That's the definition of the inverse of a function. Right? – stressed out Dec 9 '17 at 12:36
• Actually, i'm seriously embarrassed to say that i'm not able to solve this system of equations. I totally forgot how to deal with complex numbers. If you'd have some time to explain how did you solve this system i'd very much appreciate it! thank you again. – Manletta Dec 9 '17 at 12:39
• @Manletta: There is absolutely no need to be embarrassed about that. Complex numbers are definitely more confusing for calculations. In general, I wrote the system in the matrix form $A\vec{x}=\vec{b}$. Therefore, $\vec{x}=A^{-1}\vec{b}$. I calculated the inverse of the matrix $A$ and because $A$ happened to be $2\times 2$, the inverse was easy to write down. The inverse of $\pmatrix{x && y \\ z && t}=\frac{1}{xt-yz}\pmatrix{t && -y \\ -z && x}$. Note that linear algebra works over any field and $\mathbb{C}$ is a field. Therefore, everything works well. – stressed out Dec 9 '17 at 12:44
• I have one problem here, and that is that i never worked with a matrix before, we're supposed to study that in my second semester i think so i really have no clue. – Manletta Dec 9 '17 at 12:46