Show that $\left(\left\{x, -x, \frac{1}{x}, -\frac{1}{x}\right\}, \circ\right)$ is a group.

Question

Given the following problem:

Given the functions $g_1, g_2, g_3, g_4$ of $\mathbb{R}^*$ in $\mathbb{R}^*$ defined in the following way: $g_1(x) = -x$, $g_2(x) = -\frac{1}{x}$, $g_3(x) = x$ and $g_4(x) = \frac{1}{x}$. If $G = \{g_1, g_2, g_3, g_4\}$:

1. Show that $(G, \circ)$ is a group where $\circ$ is the composition of functions. Write the table.

2. Identify a generator set of $(G, \circ)$ that has the least number of elements possible.

3. Extract all the normal subgroups of $(G, \circ)$. If $H$ is one of them, describe $G$ \ $H$.

I am having lots of problems figuring out how proceed with such a set.

If I understand correctly, the composition of functions is for example:

$$ (\forall x\in\mathbb{R}^*):\enspace(g_1 \circ g_2)(x) = g_1(g_2(x)) = g_1(-\frac{1}{x}) = -(-\frac{1}{x}) = \frac{1}{x} $$

Is that so?

Then I know that to prove that $(G, \circ)$ is a group, I have to prove the following:

1. The internal law: $g_i \circ g_j \in G$.

2. Associativity: $g_i \circ (g_j \circ g_k) = (g_i \circ g_j) \circ g_k$.

3. Existence of the neutral element $g_e$ such that: $g_i \circ g_e = g_e \circ g_i = g_i$.

4. Existence of an inverse element $g_i'$ for each $g_i \in G$, we have that $g_i \circ g_i' = g_i' \circ g_i = g_e$

But how do I proceed with such a set?

Answer 1

Compute the Cayley table, taking into account that $g_3$ is the identity function:

\begin{array}{c|cccc} & g_3 & g_1 & g_2 & g_4 \\ \hline g_3 & g_3 & g_1 & g_2 & g_4 \\ g_1 & g_1 & g_3 & g_4 & g_2 \\ g_2 & g_2 & g_4 & g_3 & g_1\\ g_4 & g_4 & g_2 & g_1 & g_3 \end{array} You can notice that each element is the inverse of itself, so if this is a group it must be the Klein $4$-group $\{1,a,b,c\}$ where $a^2=1$, $b^2=1$, $c^2=1$, $ab=ba=c$, $bc=cb=a$ and $ca=ac=b$: the Cayley table is \begin{array}{c|cccc} & 1 & a & b & c \\ \hline 1 & 1 & a & b & c \\ a & a & 1 & c & b \\ b & b & c & 1 & a \\ c & c & b & a & 1 \end{array} As you see the map $g_3\mapsto 1$, $g_1\mapsto a$, $g_2\mapsto b$, $g_4\mapsto c$ is an isomorphism of “magmas”. Since the latter is a group, also the given set is a group.

Answer 2

Let's do this step by step:

1. This is just a matter of doing every possible composition. It probably can me done more elegantly, but you're being asked to write the multiplication table anyway.

2. This is true in general. Since the equality

$$ f \circ ( g \circ h) = (f \circ g) \circ h $$

is an equality of functions, one must see that it holds for every element in the domain. Thus, in this case, if $x \in \mathbb{R}^*$,

$$ (f \circ ( g \circ h))(x) = f((g\circ h)(x)) = f(g(h(x)) = (f \circ g)(h(x)) = ((f \circ g) \circ h)(x) $$

which proves the former.

3. Note that the function $e : x \in \mathbb{R}^* \mapsto x \in \mathbb{R}^*$ is the identity function on $\mathbb{R}^*$. Thus for any other such function (in particular the ones of this group),

$$ (f \circ e)(x) = f(e(x)) = f(x) = e(f(x)) = (e \circ f)(x) $$

and therefore $e \circ f = f = f \circ e$.

4. Again, when making the multiplication table, this will be a lot clearer.

As for the second and third questions, one can see that every group of $4$ elements is abelian (i.e. it is commutative), so every subgroup will be normal. Moreover, there are exactly $2$ groups of $4$ elements up to isomorphism: $\mathbb{Z}_4$ and $\mathbb{Z}_2 \oplus \mathbb{Z}_2$. The former is cyclic so it can be generated by a single element, whereas the second needs at least $2$ and that is sufficient.

Note that in your group, any function $g_i\in G$ verifies that $g_i \circ g_i = e$, so the group can't be generated by a single element (each set $\langle g_i \rangle$ would have at most 2 elements). Therefore, you'll have to look for a set of two generators. This, in particular. tells you that $G \simeq \mathbb{Z}_2 \oplus \mathbb{Z}_2$ so you will have the trivial subgroups $\{e\}$ and $G$, and three subgroups of order $2$. I'll leave you to identify these.