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This is a rather elementary question but I am teaching myself group theory and want to make sure I am getting the basics right!

I am considering how many isomorphisms there are $C_n \rightarrow C_n$. For an isomorphism, an element of some order $k$ must be mapped onto an element of order $k$. So I thought that, if $a$ was the generator in the first group, then the homomorphism could map this onto any of the elements $b^i$ in the second group were $b$ generates the second group and $i$ has to be coprime with $n$, because any of the $b^i$ with $i$ being coprime with $n$ can be thought of as the generators for the group of order $n$? Is this correct?

Secondly, I am asked how many homomorphisms there are $D_{2n}\rightarrow C_n$ Now for a homomorphism $f$, the order of $f(a)$ must divide the order of $a$. I will call the generator of order n in $D_{2n}$ r, and of order 2 $s$. And I will call a generator of order n in $C_n$ as $a$. Splitting the problem into cases according to the value of n:

  • When n=1 have trivial case that both elements $e,s \in D_{2n}$ are both in the kernel of f
  • When n=2, can choose either $f(r)=a$ and $f(s)=e$ ($e$ the ide9antity) or vice versa to satisfy the 'generator' requirement. For $n>2$, since $f(s)$ can only be order 1 or 2, it must be mapped to the identity and $f(r)=a$.

    But I am rather stuck as to what happens to the elements in $D_{2n}$ which are of the form $sr^i$ for some $i=1,2,...n-1$.

    If I consider the simple case of n=2, then I can think of $f(s\bullet r)=f(s)f(r)=ef(r)=a$ by definition for a homomorphism. Now if I consider $f(sr\bullet s)=f(sr)f(s)=f(sr)=f(r)=a$ which happens to be consistent here because $r$ is of order two and is self-inverse, so if we take $srs=ssr^{-1}=r^{-1}$ by definition for the dihedral group, this happens to be $r$ so indeed we get back $f(r)=a$ on the left hand side.

    But this does not work with any $n>2$ in general. In particular, if I assume as above that $f(s)=e$ and $f(r)=a$, then still the first case where I tried $f(s\bullet r^j)=f(s)f(r^j)=a^j$ suggesting that the $sr^j$ are mapped to $a^j$, like the $r^j$. On the other hand this suggests the second operation becomes $f(sr^j\bullet s)=f(sr^j)f(s)=f(r^j)=a^j$ but by definition for the dihedral group $sr^js=s^2r^{-j}=r^{-j}$, and clearly $f(r{-j})=a^{-j}\neq a^j$!

I am not sure if I am doing something wrong here, or perhaps there aren't any homomorphisms? I would greatly appreciate any help.

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    $\begingroup$ $D_{2n}$ has presentation $\langle r, s | r^n = e, s^2 = e, s^{-1} r s = r^{-1} \rangle$. So to give a homomorphism $D_{2n} \to G$ is equivalent to giving two elements $x, y \in G$ satisfying the same relations. In the case that $G$ is abelian (which is the case for $C_n$) then the third relation is equivalent to $x^2 = e$. $\endgroup$ – Daniel Schepler Jul 11 '17 at 23:01
  • $\begingroup$ math.stackexchange.com/questions/267880/… This should help if you're stuck on the cyclic group isomorphism part at all. $\endgroup$ – Chickenmancer Aug 29 '17 at 18:01
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1) Isomorphisms of $C_n$

We could start by counting the number of maps from $C_n$ to itself. This would be $n^n$ as each element has $n$ possible elements it can be mapped to.

Now, any homomorphism must have $\rho(e)=e$ (try and prove it). Notice that $C_n$ is cyclic and so is generated by a single element, say $a$. Now, we have $n$ possible choices for $\rho(a)$. Notice that this will determine the homomorphism as $\rho(x)^n = \rho(x^n)$. So I'm counting via a bijection from homomorphisms from $C_n$ to itself and possible values of $\rho(a)$. [If you ever study representation theory, you'll learn this result generalises in a very cool way. Also if I really wanted to be a pedant, I would make the bijection more explicit and also prove I've not over counted, but I'll leave that to you :)]

Now we have obviously overcounted for isomorphisms and we do indeed need $b:= \rho(a)$ to have order $n$. From number theory, we know that this is the same as $b = a^i$ where $i$ is coprime to $n$ (again, one ought to prove this claim to understand it fully). Thus the number of isomorphisms of $C_n$ is $\phi(n)$ where $\phi$ is the Euler Phi Function.

2) Homomorphisms from $D_{2n}$ to $C_n$

I'll use the same generators as you. Again the homomorphism will be entirely determined by its action on $r$ and $s$. We only need $\mathrm{ord}(\rho(r))|\mathrm{ord}(r)=n$ and $\mathrm{ord}(\rho(s))|\mathrm{ord}(s)=2$ and that $\rho(s^{-1}rs) = \rho(r^{-1})$ or equivalently $\rho(r)^2 = e$. The "equivalently" comes from (lengthy but basic) algebra. So $C_n$ is commutative as it is cyclic, so $\rho(s^{-1}rs) = \rho(s^{-1})\rho(r)\rho(s) = \rho(s^{-1})\rho(s)\rho(r) =\rho(r)$. But from before, $\rho(r) = \rho(r^{-1})$, so we have $\rho(r)^2 = \rho(r^{-1})\rho(r)= \rho(r^{-1}r)=\rho(e)=e$. Note that this can be summarised as $\rho(r^2) = e = \rho(s^2)$.

Consider $s$, then $\mathrm{ord}(\rho(s)) = 1$ or $\mathrm{ord}(\rho(s))=2$. The first case implies $\rho(s) = e$ and the second is only viable when $n$ is even. The same argument holds for $r$ and thus we have: $|\mathrm{Hom}(D_{2n}, C_n)| = \begin{cases} 1, & \text{if $n$ is odd} \\ 4, & \text{if $n$ is even.} \end{cases}$

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  • $\begingroup$ Hi mdave16, apologies for the late response to your post. Hope you're still able to reply... I'm unsure about a couple of things. Under the isomorphisms of Cn, you said that we know there are only n homomorphisms because $p(a^2)=p(a)^2$. Probably being a bit dim here but I don't see why this implies there can only be n homomorphisms. Secondly, for the homomorphisms from D_2n to Cn, I do not get why $p(srs)=p(r^{-1}$ is equivalent to $p(r)^2=e$? $\endgroup$ – 21joanna12 Jul 18 '17 at 17:38
  • $\begingroup$ Green tick and +1! Thank you for the swift response! $\endgroup$ – 21joanna12 Jul 18 '17 at 18:08
  • $\begingroup$ I've realised that throughout I use a number of small lemmas that one should really prove, especially the first time. I would recommend finding a book on the subject and simply writing down a list of lemmas and definitions without reading the book, and then proving them by yourself first. I think this helped me a lot for learning how to rigorously prove things. (Of course do check the actual proof to see if you have any unnecessary steps that could be avoided or if you have some necessary steps which you did avoid, find out why they are necessary or unnecessary.) $\endgroup$ – mdave16 Jul 19 '17 at 6:50

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