# Klein 4-group

## 1 Klein 4-group as a symmetry group

###### Proposition 1.

$V$

###### Proof.

Suppose $V$ is the automorphism group of a simple graph $G$. Because $V$ contains the permutations $(12)(34)$, $(13)(24)$ and $(14)(23)$ it follows the degree of every vertex is the same – we can map every vertex to every other. So $G$ is a regular graph  on 4 vertices. This makes $G$ isomorphic to one of the following 4 graphs:

 $\xy<10mm,0mm>:<0mm,10mm>::(0,0)*+{1}="1";(1,0)*+{2}="2";(1,1)*+{3}="3";(0,1)*+% {4}="4";\qquad\xy<10mm,0mm>:<0mm,10mm>::(0,0)*+{1}="1";(1,0)*+{2}="2";(1,1)*+{% 3}="3";(0,1)*+{4}="4";"1";"2"**@{-};"3";"4"**@{-};\qquad\xy<10mm,0mm>:<0mm,10% mm>::(0,0)*+{1}="1";(1,0)*+{2}="2";(1,1)*+{3}="3";(0,1)*+{4}="4";"1";"2"**@{-}% ;"2";"3"**@{-};"3";"4"**@{-};"4";"1"**@{-};\qquad\xy<10mm,0mm>:<0mm,10mm>::(0,% 0)*+{1}="1";(1,0)*+{2}="2";(-0.5,0.86)*+{3}="3";(-0.5,-0.86)*+{4}="4";"1";"2"*% *@{-};"1";"3"**@{-};"1";"4"**@{-};"2";"3"**@{-};"2";"4"**@{-};"3";"4"**@{-};.$

In order the automorphism groups of these graphs are $S_{4}$, $\langle(12),(34)\rangle$, $\langle(12),(1234)\rangle$ and $S_{4}$. None of these are $V$, though the second is isomorphic to $V$. ∎

 $\xy<10mm,0mm>:<0mm,10mm>::(0,0)*+{1}="1";(2,0)*+{2}="2";(2,1)*+{3}="3";(0,1)*+% {4}="4";"1";"2"**@{-};"2";"3"**@{-};"3";"4"**@{-};"4";"1"**@{-};$

We can rotate by $180^{\circ}$ which corresponds to the permutation $(13)(24)$. We can also flip the rectangle over the horizontal diagonal which gives the permutation $(14)(23)$, and finally also over the vertical diagonal which gives the permutation $(12)(34)$.

 $\xy<10mm,0mm>:<0mm,10mm>::(0,0)*+{3}="1";(2,0)*+{4}="2";(2,1)*+{1}="3";(0,1)*+% {2}="4";"1";"2"**@{-};"2";"3"**@{-};"3";"4"**@{-};"4";"1"**@{-};,\quad\xy<10mm% ,0mm>:<0mm,10mm>::(0,0)*+{4}="1";(2,0)*+{3}="2";(2,1)*+{2}="3";(0,1)*+{1}="4";% "1";"2"**@{-};"2";"3"**@{-};"3";"4"**@{-};"4";"1"**@{-};,\quad\xy<10mm,0mm>:<0% mm,10mm>::(0,0)*+{2}="1";(2,0)*+{1}="2";(2,1)*+{4}="3";(0,1)*+{3}="4";"1";"2"*% *@{-};"2";"3"**@{-};"3";"4"**@{-};"4";"1"**@{-};.$

An important corollary to this realization is

###### Proposition 2.

Given a square with vertices labeled in any way by $\{1,2,3,4\}$, then the full symmetry group (the dihedral group  of order 8, $D_{8}$) contains $V$.

## 2 Klein 4-group as a vector space

As $V$ is isomorphic to $\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$ it is a 2-dimensional vector space over the Galois field $\mathbb{Z}_{2}$. The projective geometry of $V$ – equivalently, the lattice of subgroups – is given in the following Hasse diagam:

 $\xy<10mm,0mm>:<0mm,10mm>::(0,0)*+{\langle()\rangle}="1.1";(-2,1)*+{\langle(12)% (34)\rangle}="2.1";(0,1)*+{\langle(13)(24)\rangle}="3.1";(2,1)*+{\langle(14)(2% 3)\rangle}="4.1";(0,2)*+{V}="5.1";"2.1";"1.1"**@{-};"3.1";"1.1"**@{-};"4.1";"1% .1"**@{-};"5.1";"4.1"**@{-};"5.1";"3.1"**@{-};"5.1";"2.1"**@{-};$

The automorphism group of a vector space is called the general linear group  and so in our context $\operatorname{Aut}V\cong GL(2,2)$. As we can interchange any basis of a vector space we can label the elements $e_{1}=(12)(34)$, $e_{2}=(13)(24)$ and $e_{3}=(14)(23)$ so that we have the permutations $(e_{1},e_{2})$ and $(e_{2},e_{3})$ and so we generate all permutations on $\{e_{1},e_{2},e_{3}\}$. This proves:

###### Proposition 3.

$\operatorname{Aut}V\cong GL(2,2)\cong S_{3}$. Furthermore, the affine linear group of $V$ is $AGL(2,2)=V\rtimes S_{3}$.

## 3 Klein 4-group as a normal subgroup

Because $V$ is a subgroup of $S_{4}$ we can consider its conjugates. Because conjugation in $S_{4}$ respects the cycle structure  . From this we see that the conjugacy class  in $S_{4}$ of every element of $V$ lies again in $V$. Thus $V$ is normal. This now allows us to combine both of the previous sections       to outline the exceptional nature (amongst $S_{n}$ families) of $S_{4}$. We collect these into

###### Theorem 4.
1. 1.

$V$$S_{4}$.

2. 2.

$V$ is contained in $A_{4}$ and so it is a normal subgroup of $A_{4}$.

3. 3.

$V$ is the Sylow 2-subgroup of $A_{4}$.

4. 4.

$V$$S_{4}$, that is, the $2$-core of $S_{4}$.

5. 5.

$S_{4}/V\cong S_{3}$.

6. 6.

$S_{4}\cong AGL(2,2)\cong V\rtimes S_{3}$.

###### Proof.

We have already argued that $V$ is normal in $S_{4}$. Upon inspecting the elements of $V$ we see $V$ contains only even permutations  so $V\leq A_{4}$ and consequently $V$ is normal in $A_{4}$ as well. As $|A_{4}|=12$ and $|V|=4$ we establish $V$ is a Sylow 2-subgroup of $A_{4}$. But $V$ is normal so it the Sylow 2-subgroup of $A_{4}$ (Sylow subgroups are conjugate.)

Now notice that the dihedral group $D_{8}$ acts on a square and so it is represented as a permutation group  on 4 vertices, so $D_{8}$ embeds in $S_{4}$. As $|D_{8}|=8$ and $|S_{4}|=24$, $D_{8}$ is a Sylow 2-subgroup of $S_{4}$ and so all Sylow 2-subgroups of $S_{4}$ are embeddings  of $D_{8}$ (in particular various relabellings of the vertices of the square.) But by Proposition  2 we know that each embedding contains $V$. As there are 3 non-equal embeddings of $D_{8}$ (think of the 3 non-equal labellings of a square) we know that the intersection of these $D_{8}$ is a proper subgroup  of $D_{8}$. As $V$ is a maximal subgroup of each $D_{8}$ and contained in each, $V$ is the intersection of all these embeddings.

Now the action of $S_{4}$ by conjugation on the Sylow 2-subgroups $D_{8}$ permutes all 3 (again Sylow subgroups are conjugate) so $S_{4}\mapsto S_{3}$. Indeed, $V$ is in the kernel of this action as $V$ is in each $D_{8}$. Indeed a three cycle $(123)$ permutes the $D_{8}$’s with no fixed point  (consider the relabellings) and $(12)$ fixes only one. So $S_{4}$ maps onto $S_{3}$ and so the kernel is precisely $V$. Thus $S_{4}/V=S_{3}$.

Now we can embed $S_{3}$ into $S_{4}$ as $\langle(123),(12)\rangle$ so $V\cap S_{3}=1$, $VS_{3}=S_{4}$ so $S_{4}=V\ltimes S_{3}$. Finally, $AGL(2,2)$ acts transitively on the four points of the vector space $V$ so $AGL(2,2)$ embeds in $S_{4}$. And by Proposition 3 we conclude $S_{4}\cong AGL(2,2)$. ∎

We can make similar   arguments about subgroups of symmetries for larger regular polygons  . Likewise for other 2-dimensional vector spaces we can establish similar structural properties. However it is only when we study we involve $V$ that we find these two methods intersect in a this exceptionally parallel   fashion. Thus we establish the exceptional structure of $S_{4}$. For all other $S_{n}$’s, $A_{n}$ is the only proper normal subgroup.

We can view the properties of our theorem in a geometric way as follows: $S_{4}$ is the group of symmetries of a tetrahedron   . There is an induced action of $S_{4}$ on the six edges of the tetrahedron. Observing that this action preserves incidence relations  one gets an action of $S_{4}$ on the three pairs of opposite edges.

## 4 Other properties

$V$ is non-cyclic and of smallest possible order with this property.

$V$ is the symmetry group of the Riemannian curvature tensor.

 Title Klein 4-group Canonical name Klein4group Date of creation 2013-03-22 12:49:02 Last modified on 2013-03-22 12:49:02 Owner Algeboy (12884) Last modified by Algeboy (12884) Numerical id 26 Author Algeboy (12884) Entry type Topic Classification msc 20K99 Synonym Klein four-group Synonym Viergruppe Related topic GroupsInField Related topic Klein4Ring Related topic PrimeResidueClass Related topic AbelianGroup2