abstract-algebrafinite-groupsgroup-theory
Klein 4-group is a symmetry group of rectangle (or rombus).
And as far as I understand it is not isomorphic to Dihedral group of order 4.
Because Dihedral group of order 4 is a group of rotations of square.
If this is true then why Wiki says that they are isomorphic:
The Klein four-group is the smallest non-cyclic group. It is however an >abelian group, and isomorphic to the dihedral group of order (cardinality) 4
What am I missing here?
Just a “visual” construction of the isomorphism between $\mathfrak S(4)/V_4$ and $\mathfrak S(3)$...
It's quite well known that $4 = 2 +2$. Concretely, it means that if you have four things, you can always group them into two pairs. What's remarkable is that there are exactly three such groupings: $$ \{1, 2, 3, 4\} = \{1, 2\} \sqcup \{3,4\} = \{1, 3\} \sqcup \{2,4\} = \{1, 4\} \sqcup \{2,3\}.$$
If I call these three groupings $\mathbf{Order}$, $\mathbf{Parity}$ and $\mathbf{5sum}$, every permutation of $\{1,2,3,4\}$ induces a permutation of these three things. For example, the $4$-cycle $(1234)$ induces the transposition $(\mathbf{Order}\ \mathbf{5sum})$. So I get a homomorphism $$ \mathfrak S(4) \to \mathfrak S(\{\mathbf{Order}, \mathbf{Parity}, \mathbf{5sum}\})\simeq \mathfrak S(3).$$ It's quite easy to show that this morphism is surjective (we already found the transposition $(\mathbf{Order}\ \mathbf{5sum})$ in its image, it's not hard to find another transposition or a $3$-cycle, and that will be enough to generate the whole of $\mathfrak S(3)$.)
Let's give a proof that the kernel is exactly $V_4$: let $\sigma$ be a nontrivial permutation in it. So for example, there are two elements $a \neq b$ such that $\sigma(a) = b$ (let's call $c$ and $d$ the two remaining elements) In that case, because $\sigma$ preserves the grouping $\{a,b\} \sqcup \{c,d\}$, it must send $b$ back to $a$. And because it preserves $\{a,c\} \sqcup \{b,d\}$ and it exchanges $a$ and $b$, it must exchange $c$ and $d$ as well. So $\sigma = (a\, b)(c\, d) \in V_4$.
Therefore, the factorisation theorem gives you an isomorphism $\mathfrak S(4)/V_4 \to \mathfrak S(3)$.
Of course, this proof is not the shortest (well, it certainly isn't the shortest to write, but if you're allowed to make a lot of pictures or to play with four actual tokens, essentially all the arguments become self-evident). But it really makes the isomorphism concrete. There's an isomorphism between $\mathfrak S(4)/V_4$ and $\mathfrak S(3)$ because $4$ equals $2+2$ in $3$ different ways...
A final remark: if $n \neq 4$, the only proper normal subgroup of $\mathfrak S(n)$ is $\mathfrak A(n)$. That means that the situation I described here is pretty exceptional. If you want $\mathfrak S(n)$ to act on fewer than $n$ objects, the only interesting actions are :
I may be overenthusiastic, but this simple remark gives much cachet to this innocent-looking isomorphism. (Another peculiarity of the same kind is that the only interesting action of $\mathfrak S(n)$ on $n+1$ tokens is an action of $\mathfrak S(5)$ on 6 tokens, coming from another exceptional behaviour of the symmetric group.)
Note that $20 = 2^2 \cdot 5$.
By Fundamental theorem of Finitely Generated Abelian Group, there are two distinct abelian groups of order $20$: $\mathbb{Z}_{20}$ and $\mathbb{Z}_{10} \times \mathbb{Z}_2$.
Now let $G$ be a nonabelian group of order $20$. By Sylow’s Theorem, $n_5 = 1$, so that $G$ has a unique (hence normal) Sylow $5$-subgroup $H \cong \mathbb{Z}_5$. Now let $K$ be any Sylow $2$-subgroup of $G$. By Lagrange, we have $H \cap K = 1$, so that $G = HK$. By the recognition theorem for semidirect products, $G \cong H \rtimes_\varphi K$ for some $\varphi : K \rightarrow \mathsf{Aut}(H)$. Evidently, classifying the nonabelian groups of order $20$ is equivalent to determining the nonisomorphic groups constructed in this manner. To that end, let $H = \mathbb{Z}_5 = \langle y \rangle$. Note that $\mathsf{Aut}(H) = \langle \alpha \rangle \cong \mathbb{Z}_4$; where $\alpha(y) = y^2$.
Let $K = \mathbb{Z}_4 = \langle x \rangle$. There are four distinct homomorphisms $K \rightarrow \mathsf{Aut}(H)$.
If $\varphi_1(x) = 1$, then $\varphi_1$ is trivial; this contradicts the nonabelianicity of $G$.
If $\varphi_2(x) = \alpha$, then $\mathbb{Z}_5 \rtimes_{\varphi_2} \mathbb{Z}_4 $is indeed a nonabelian group of order $20$.
If $\varphi_3(x) = \alpha^2$, then $\mathbb{Z}_5 \rtimes_{\varphi_3} \mathbb{Z}_4$ is indeed a nonabelian group of order $20$. Moreover, since $\mathsf{ker}\ \varphi_3 \cong \mathbb{Z}_2$ and $\mathsf{ker}\ \varphi_2 \cong 1$,$ H \rtimes_{\varphi_3} K \not\cong H \rtimes_{\varphi_2} K$.
If $\varphi_4(x) = \alpha^3$, then $\mathsf{im}\ \varphi_4 = \mathsf{im}\ \varphi_2$. Since $\mathbb{Z}_4$ is cyclic, by a previous theorem, $H \rtimes_{\varphi_4} K \cong H \rtimes_{\varphi_2} K$.
Thus there are two distinct groups of order 20 which have a cyclic Sylow 2-subgroup.
Suppose now that $K = \mathbb{Z}_2^2 = \langle a \rangle \times \langle b \rangle$. Again, $\psi : \mathbb{Z}_2^2 \rightarrow \mathbb{Z}_4$ is determined uniquely by $\psi(a)$ and $\psi(b)$, and is indeed a homomorphism provided $|\psi(a)|$ and $|\psi(b)|$ divide $2$. We thus have $\psi(a)$, $\psi(b) \in \{ 1, \alpha^2 \}$, for a total of four choices.
If $\psi_1(a) = \psi_1(b) = 1$, then $\psi_1 = 1$, contradicting the nonabelianicity of $G$.
If $\psi_2(a) = \alpha^2$ and $\psi_2(b) = 1$, then $\mathbb{Z}_5 \rtimes_{\psi_2} \mathbb{Z}_2^2$ is indeed a nonabelian group of order $20$.
If $\psi_3(a) = 1$ and $\psi_3(b) = \alpha^2$, then $\varphi_3 = \varphi_2 \circ \theta$, where $\theta(a) = b$ and $\theta(b) = a$. Clearly $\theta$ is an automorphism of $\mathbb{Z}_2^2$. By a lemma to a previous theorem, we have $H \rtimes_{\psi_3} K \cong H \rtimes_{\psi_2} K$.
If $\psi_4(a) = \alpha^2$ and $\psi_4(b) = \alpha^2$, then $\psi_4 = \psi_2 \circ \theta$, where $\theta(a) = a$ and $\theta(b) = ab$. Clearly $\theta$ is an automorphism of $\mathbb{Z}_2^2$. By a lemma to a previous theorem, we have $H \rtimes_{\psi_4} K \cong H \rtimes_{\psi_2} K$.
Thus there is a unique nonabelian group of order 20 which has an elementary abelian Sylow 2-subgroup.
In summary, the distinct groups of order $20$ are as follows. We let $\mathbb{Z}_5 = \langle y \rangle$, $\mathbb{Z}_4 = \langle x \rangle$, and $\mathbb{Z}_2^2 = \langle a \rangle \times \langle b \rangle$.
$Z_{20}$,
$Z_{10} \times Z_2$,
$Z_5 \rtimes_{\varphi_3} Z_4$, where $\varphi_3(x)(y) = y^{-1}$.
$Z_5 \rtimes_{\varphi_2} Z_4$, where $\varphi_2(x)(y) = y^2$
$Z_5 \rtimes_\psi Z_2^2$, where $\psi(a)(y) = y^{-1}$ and $\psi(b)(y) = y$.
(Source: Crazyproject)
Best Answer
You are confusing 'the dihedral group with 4 elements' with 'the dihedral group that is the symmetry group of the square'. The dihedral group with 4 elements is the set of symmetries of a bigon (i.e. a 'polygon' with 2 vertices, which can be thought of as two curved lines joined together) or of a rectangle (thanks to Akiva Weinberger).
The dihedral group that is the symmetry group of a square is of order 8 and usually is denoted $D_4$.