# Showing that the Klein-four group is associative without checking 64 options

Showing that the Klein-four group is associative without checking 64 options

I don't know how to go about this. I could use the fact it's abelian to eliminate a few duplicates, but many options remain. Is there an easier way of proving that the Klein-four is associative?

• It's a direct product of groups $\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$ Commented Jan 11, 2021 at 18:08
• It depends on how you’ve defined it. A standard thing to do here is to prove once and for all that function composition is associative. This implies that all of the symmetric groups are associative, and then you can exhibit the Klein four group as a group of permutations. Commented Jan 11, 2021 at 18:28

## 3 Answers

It's a Monday afternoon, so let's have some fun:

Let's start with the assumption that there is a set $$S = \{e, a, b, c\}$$ with binary operation $$*$$ having Cayley table: $$\begin{array} {|r|r|}\hline \cdot & e & a & b & c \\ \hline e & e & a & b & c \\ \hline a & a & e & c & b \\ \hline b & b & c & e & a \\ \hline c & c & b & a & e \\ \hline \end{array}$$

We're going to use Light's associativity test. Because $$S$$ is generated by $$a$$ and $$b$$ (i.e. $$e = a * a$$ and $$c = a * b$$), it suffices to show that for all $$x, y \in S$$ we have both $$x * (a * y) = (x * a) * y$$, and $$x * (b * y) = (x * b) * y$$.

Let's check the first of these claims: start by defining binary operations $$x \star y = x * (a * y)$$ and $$x \circ y = (x * a) * y$$. We'll be down when we show that the Cayley tables for $$\star$$ and $$\circ$$ agree. There are a few optimizations we can take based on the circumstances:

• As the Cayley table is symmetric about the diagonal, we know that $$(S, *)$$ is abelian. It follows that $$x \star y = y \circ x$$ (write it out!). Therefore $$x \star x = x \circ x$$. Moreover, if $$x \star y = x \circ y$$ then also $$y \star x = y \circ x$$ (again, convince yourself!).

• We don't need to include the identity element in our table, because if either $$x = e$$ or $$y = e$$, then $$x \star y = x \circ y$$.

So now the question is how much of the two Cayley tables do we have to check? Using the above remarks, you get down to something like the below, where "No" means "No, you don't need to check it": $$\begin{array} {|r|r|}\hline \star & e & a & b & c \\ \hline e & No & No & No & No \\ \hline a & No & No & ? & ? \\ \hline b & No & No & No & ? \\ \hline c & No & No & No & No \\ \hline \end{array}$$

When the smoke clears, you see that it suffices to check: $$a \star b = a \circ b,\ a \star c = a \circ c,\ \text{and } b \star c = b \circ c.$$

To check each equation, you need to do 4 multiplications (i.e. Cayley table look-ups). For instance, \begin{align*} a \star b &= a * (a * b) = a * c = b,\\ a \circ b &= (a * a) * b = e * b = b. \end{align*} To conclude: to check whether the Klein four-group is associative, for each of its 2 generators, it suffices to check 3 "associativity equations" in $$(S, *)$$.

I'm curious whether there are further simplifications one can make by using algebraic properties specific to the Klein four-group table, but this is all the mileage I can get out of Light's test alone.

Using the $$K_4$$ properties:
: elements are self-inverse
: operation is commutative

Consider $$x,y,z \in K_4$$, not necessarily distinct. We ask if $$x\cdot (y \cdot z) = (x\cdot y) \cdot z$$.

If any of $$x,y,z$$ is the identity, the equality is immediately true.

If $$x,y,z$$ are distinct, the result is the identity (since inverse is unique):
$$x\cdot (y \cdot z) = x\cdot x = e$$
$$(x\cdot y) \cdot z = z\cdot z = e$$

For $$x=z$$, commutativity gives:
$$(x\cdot y)\cdot x =(y\cdot x)\cdot x =x\cdot (y\cdot x )$$

If $$y=x\neq z$$,
$$(x\cdot y) \cdot z = e\cdot z = z$$ and
$$x\cdot (y \cdot z) = y\cdot w = z$$ where $$w$$ is the other element, $$w\neq (x=y),z$$
and similarly if $$y=z\neq x$$.

• About the last part, couldn't $w$ be equal to e? then $y*w=y$ and the equality becomes false? And also, what about the instance in which $y\neq x\neq z$? Commented Jan 12, 2021 at 7:57
• Elements are self-inverse, so $y\cdot z$ couldn't be $e$ (or $y$ or $z$). $x,y,z$ distinct is covered in the second point. Commented Jan 12, 2021 at 8:20
• But it seems as if $w\neq y,z,e$, then $w=x$, but then $y*w=y*x=x*x=e\neq z$, and it falsifies the equality. Commented Jan 12, 2021 at 8:27
• In that section of the argument $x$ and $y$ are the same (non-identity) element, $z$ is a second distinct one, and I have co-opted $w$ to stand in for the third. The actual elements are generally called $a,b,c$. Commented Jan 12, 2021 at 8:49

Thank you all for your help! I will post here my final answer:

First, we'll write down the Cayley Table: $$\begin{array} {|r|r|}\hline * & e & a & b & c \\ \hline e & e & a & b & c \\ \hline a & a & e & c & b \\ \hline b & b & c & e & a \\ \hline c & c & b & a & e \\ \hline \end{array}$$

From the symmetry along the main diagonal of the table, it's immediate that the operation is Abelian. We'll check whether $$x,y,z\in P$$ satisfy $$(x*y)*z=x*(y*z)$$.

If any of the elements are $$e$$, the equality is immediately true.

If $$x=z$$ the equality is immediately true (due to the operation being Abelian).

If $$y=x\neq z$$, the left side of the equation is $$(x*y)*z=e*z=z$$. For the right side, we'll define $$w=y*z$$. So $$x*(y*z)=y*w$$. By definition, $$w$$ is the third distinct element so that $$w\neq e$$. (If $$x=y=a$$ and $$z=c$$ then $$w=b$$). Hence $$y*w=z$$.

Similarly, it can be proven for $$y=z\neq x$$.

for $$x\neq y\neq z$$, $$(x*y)*z=z*z=e$$ and $$x*(y*z)=x*x=e$$.

So the equality is true for any $$x,y,z$$, the operation is associative.