I'm trying to do these three exercises for my math study:

a) Determine the number of homomorphisms from $D_5$ to $\mathbb{R^*}$
b) Determine the number of homomorphisms from $D_5$ to $S_4$
c) give an injective homomorphism form $D_5$ to $S_5$

I think I solved a):

The amount of homomorphisms is 1. The group $D_5$ is generated by a rotations and reflections. every reflection has order 2, and the rotations have order 1(the rotation about nothing) and order 5(the 4 rotations left). so we have to sent the reflection to -1 or 1, because the order of $f(x)$ has to divide the order of $x$ and the rotation about 0 degrees has only 1 as option to send it to. Combining that gives us only 1 homomorphism.

For b), I think the answer is 6! = 720, because every reflection in $D_5$ has to go to a 2-cycle in $S_4$, because the 2-cycles generate $S_4$. There are six of them, so the number of different homomorphisms are 6!.

Can you tell me if this is correct, and explain c) to me?

Thanks in advance!

  • $\begingroup$ For c), imagine a pentagon with the vertices at the root of $z^5=1$ on complex plane. Let's assign $1$ to the vertex on $z=1$ and assign the next integer to the next vertex counterclockwise. Then, the reflection about real line corresponds to $(2 5)(3 4)$ in $S_5$. The rotation corresponds to $(1 2 3 4 5)$. These two elements in $S_5$ generate a subgroup of $S_5$, so this is an injective homomorphism. $\endgroup$ – Math.StackExchange Mar 30 '15 at 11:26
  • $\begingroup$ for b) you assume you can choose the images of all reflections independently, that is not true, actually if $f:D_5\rightarrow S_5$ is a morphism and $s$, $t$ two reflections then $f(s)=f(t)$. $\endgroup$ – Clément Guérin Mar 30 '15 at 11:30
  • $\begingroup$ @Clément Guérin And why is that? $\endgroup$ – Peter Mar 30 '15 at 11:31
  • $\begingroup$ Well take $r$ a rotation and $s$ a reflection generating $D_5$. Now if $t$ is another reflection then (from a carefull study of $D_5$) you must have $t=r^ks$, then $f(t)=f(r)^kf(s)$ but $f(r)$ is an element of order dividing $5$ in $S_4$ and so is the identity, from this you have $f(t)=f(s)$. I use this (without saying it explicitely) in my answer. $\endgroup$ – Clément Guérin Mar 30 '15 at 11:34

For a) the answer is two because :

$$Hom(D_5,\mathbb{R}^*)=Hom(D_5/[D_5,D_5],\mathbb{R}^*)=Hom(Z/2Z,\mathbb{R}^*) $$

The cardinal of the last one is $2$.

Edit : another way without abelianization.

Take $r$ a rotation and $s$ a reflection generating $D_5$. Take $f\in Hom(D_5,\mathbb{R}^*)$. You have $f(r)^{5}=1$, the only real number verifying this is $1$.

Now every element of $D_5$ is written as $r^k$ or $r^ks$. The image of $r^k$ by $f$ must be $1$. The image of $r^ks$ by $f$ is $f(r^ks)=f(r)^kf(s)=f(s)$.

We then see that the morphism $f$ is only determined by its value in $s$. This value must verify $f(s)^2=1$ so it is either $-1$ or $1$. This shows that there are at most $2$ such morphisms. To justify that there exist a non-trivial morphism from $D_5$ to $\mathbb{r}^*$ one can find a non-trivial morphism from $D_5$ to $Z/2Z=\{\pm 1\}$ by :

$$D_5\rightarrow D_5/<r> $$

For b) you see that if $f:D_5\rightarrow S_4$ is a morphism then $f(r)=Id$ (because there are no element of order $5$). Now such morphisms have then image in a subgroup of order $2$ of $S_4$. It follows easily that the number of such morphisms is the number of elements of order dividing $2$ in $S_4$, you have :

$$\frac{4\times 3}{2}=6\text{ transpositions and } 3\text{ double transpositions and the identity.} $$

you then get $10$ morphisms from $D_5$ to $S_4$.

Now for c), the idea is to find a subgroup $H$ of $S_5$ isomorphic to $D_5$ (then the isomorphism from $D_5$ to $H$ gives you the injective homomorphism from $D_5$ to $H$).

It suffices (because $D_5=Z/5Z\rtimes_{-1} Z/2Z$) to exhibit an element $\tau\in S_5$ such that :

$$\tau^2=Id\text{ and } \tau(1,2,3,4,5)\tau^{-1}=(1,2,3,4,5)^{-1}=(1,5,4,3,2) $$

Then you can verify that :

$$<(1,2,3,4,5)>\rtimes_{conj} <(2,5)(3,4)> $$

Is a subgroup of $S_5$ isomorphic to $D_5$.

  • $\begingroup$ I don't understand what you mean by a). Is there an easier way to explain it? $\endgroup$ – Peter Mar 30 '15 at 11:35
  • $\begingroup$ Yes there is a better way, I'll edit directly my answer. $\endgroup$ – Clément Guérin Mar 30 '15 at 11:36

For part c) you can take the generators $r=(12345)$ and $s=(25)(34)$ of $D_5$ and map them via the identity to $S_5$. This yields an injective homomorphism $\phi\colon D_5\rightarrow S_5$. (Note that $srs=s^{-1}$).


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