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?

  • 3
    $\begingroup$ I think you should limit this question to proving only $(1)$, listed in the highlighted problem statement. Also, you are correct about your understanding of the composition of two functions. I changed some wording to indicate the English version of, e.g., asociatividad, which is associativity And the word you needed in the fourth group axiom needs to be "inverse element". You define the inverse element very well. $\endgroup$
    – amWhy
    Jul 8, 2018 at 21:19
  • $\begingroup$ @amWhy Thanks for the corrections. Some words got lost in my translation. As it relates to focusing on $(1)$ of the problem, any specific reason? $\endgroup$ Jul 8, 2018 at 21:26
  • 1
    $\begingroup$ Since you have only four elements, it is reasonable to proceed as you did when checking your understanding of composition of functions. E.g., you have already shown that $g_1 \circ g_2 = g_1(g_2(x)) = g_1(-\frac{1}{x}) = -(-\frac{1}{x}) = \frac{1}{x} = g_4\in G$. The composition of each function with another needs to be one of four elements in G. Recall that for function composition, given functions $f, h$, it is not always true that $f\circ h = h\circ f$, so it would be good to check out $g_1 \circ g_2$ as well. $\endgroup$
    – amWhy
    Jul 8, 2018 at 21:28
  • 1
    $\begingroup$ @amWhy so should I edit the question and remove the others? $\endgroup$ Jul 8, 2018 at 21:33
  • 1
    $\begingroup$ That's what I would suggest. Else you might get just quick suggestions for each point, in an answer, where I think you might find it more helpful to get thorough help for each section. And I think you can be working on your table that you need, as you confirm that the group is closed under function composition. (the internal law.) Of course, if you feel the answer before answers all your questions, then you are free to keep it as is. I was simply suggesting what I thought might be more helpful to you. $\endgroup$
    – amWhy
    Jul 8, 2018 at 21:36

2 Answers 2


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.


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.

  1. 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 $e \circ f = f = f \circ e$.

  1. 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.

  • 1
    $\begingroup$ Note that wrt your last paragraph, $\langle g_3\rangle = \{g_e\}$ with only one element, where $g_3 = g_e = x$ is the identity function, hence has order one. So not all elements of $G$ generate a subgoup of order two, as you suggest. $\endgroup$
    – amWhy
    Jul 8, 2018 at 22:19
  • $\begingroup$ @amWhy sorry, I meant to say 'at most'. Fixed. $\endgroup$
    – qualcuno
    Jul 8, 2018 at 22:32
  • $\begingroup$ No problem, Guida A.! (I'm pretty prolific in typing typos, or leaving out one or two word, even though I'm thinking them!) $\endgroup$
    – amWhy
    Jul 8, 2018 at 22:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .