How many homomorphisms are there from $D_5$ to $V_4$?

Question

Question: How many homomorphisms are there from $D_5$ to $V_4$, where $D_5$ is the dihedral group of order $10$ and $V_4$ the Klein four-group? I've used the fact that since $V_4$ is abelian, the commutator subgroup of $D_5$ is contained in the kernel of any homomorphism. However, I am having trouble determining the order of the commutator subgroup.

Thanks in advance :)

Answer 1

The commutator of $D_5$ has five elements. But we can also proceed without having this information.

Let $r,s$ be the rotation and the spiegelung (reflection) generating $D_5$, seen as the symmetry group of the pentagram. There is a relation joining $r,s$. But $r^5=1$, so the image of $r$ is the one in $V_4$, the only element of order dividing $5$ in $V_4$. It remains to see that $s$ can be mapped into each element of $V_4$.

So there are four homomorphisms $D_5\to V_4$.

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Computer aid, here sage:

sage: D5 = DihedralGroup(5)
sage: D5.order()
10
sage: V4 = KleinFourGroup()
sage: D5.gens()
[(1,2,3,4,5), (1,5)(2,4)]
sage: D5.commutator()
Permutation Group with generators [(1,3,5,2,4)]

sage: r, s = D5.gens()
sage: print "r = %s of order %s" % (r, r.order())
r = (1,2,3,4,5) of order 5
sage: print "s = %s of order %s" % (s, s.order())
s = (1,5)(2,4) of order 2
sage: for v in V4:
....:     f = D5.hom( [ V4.one(), v ], V4 )
....:     print f
....:     print "f(r) = %s and f(s) = %s" % (f(r), f(s))
....:     
Group morphism:
  From: Dihedral group of order 10 as a permutation group
  To:   The Klein 4 group of order 4, as a permutation group
f(r) = () and f(s) = ()
Group morphism:
  From: Dihedral group of order 10 as a permutation group
  To:   The Klein 4 group of order 4, as a permutation group
f(r) = () and f(s) = (3,4)
Group morphism:
  From: Dihedral group of order 10 as a permutation group
  To:   The Klein 4 group of order 4, as a permutation group
f(r) = () and f(s) = (1,2)
Group morphism:
  From: Dihedral group of order 10 as a permutation group
  To:   The Klein 4 group of order 4, as a permutation group
f(r) = () and f(s) = (1,2)(3,4)

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Final later EDIT:

Above there is a computation of the commutator group for $D_5$ done by sage. Let us do this manually. First of all, let us write explicitly the relation of the two generators of $D_5$, $r$ (geometrically interpreted as) the rotation of the plane around origin with angle $2\pi/5$, and $s$ the symmetry, (spiegelung, i.e.) the reflection w.r.t. the $Ox$ axis. Imagine also the points $\zeta_k=\exp \frac{2\pi i}5$, $k=0,1,2,3,4$, in the same complex plane. Then $$ srs^{-1}=srs=r^{-1}\ , $$ and this relation can be visualized as follows. Consider the complex plane, pass via $s$ in the mirrored world, the mirrored rotation $r$ is then in the opposite direction, it is the way the mirror works. It follows $$ srs^{-1}r^{-1}=r^{-2}\ . $$ The subgroup generated by $r^{-2}$ is ths subgroup of rotations, $R=\{1,r,r^2,r^3, r^4\}$. To see that this is the commutator group, we use the non-trivial "determinant" morphism, which maps "a symmetry" in $D_5$ to $+1$, respectively $-1$, depending on the conservation of the orientation of plane. By geometric means it is well defined. Commutators are mapped to $+1$, since the image group is commutative. And exactly the rotation elements in $R$ are mapped to $+1$. No other elements can be in the commutator.

Answer 2

How many homomorphisms are there from $D_5$ to $V_4$?

The only normal subgroups of $D_5$ are $\{e\}$ , $D_5$ itself, the subgroup $H$ of all rotations. So by first isomorphism theorem, if $f$ is such a homomorphism, then $D_5/K \simeq Im(D_5)$ where $K$ is a kernel of a homomorphism. Here obviously $K \neq \{e\}$. So $K$ may be either $D_5$ or $H$. In the first case, $f$ is trivial. In the second case, $f$ is $$f: x\mapsto \begin{cases} e & \text{if}\; x\; \text{is a rotation}\\a & \text{if}\; x\; \text{is a reflection}\end{cases}$$ where $a \in V_4$ is of order $2$

Answer 3

Since $2\nmid 5$, all the four elements of order $5$ must be sent to $1$. As for the five elements of order $2$, $\varphi(sr^k)=$ $\varphi(s)\varphi(r^k)=$ $\varphi(s)$. Therefore, for $V_4=\{1,a,b,ab\}$: \begin{alignat}{1} &s\mapsto a \\ &s\mapsto b \\ &s\mapsto ab \\ \end{alignat} completely define the only three nontrivial homomorphisms.

Answer 4

The commutator of $D_5$ is $\langle r\rangle \cong \Bbb Z_5.$

As a result it's in the kernel (there's two ways to see that: $V_4$ has order $4,$ and it's abelian) , and the homomorphisms have every reflection, $$r^ks,1\le k\le5$$ going to the same element of order $2$.

So there's $3,$ $4$ including the trivial one.