Help on homomorphisms from $C_{12}$ to $C_2 \times D_5$.

Question

I'm taking an abstract algebra course and ran into a problem classifying the possible homomorphisms $\phi: C_{12} \rightarrow C_2 \times D_5$. We were asked to find the possible homomorphisms explicitly.

Now given $\phi: G \rightarrow H$, we know |$im(\phi)$| $*$ |$ker(\phi)$| = |$G$|, and that |$im(\phi)$| must divide |$H$|. This led to the conclusion that |$im(\phi)$| $= 1, 2,$ or $4$.

Then, looking at the problem specifically, knowing $C_{12} =$ <$a$>, with |$a$| = 12, we can classify the elements of $C_{12}$ by their order. Namely, we have $a^1, a^5, a^7, a^{11}$ of order 12, $a^2, a^{10}$ of order 6, $a^3, a^9$ of order 4, $a^4, a^8$ of order 3, and $a^6$ of order 2.

In addition, $C_2 \times D_5$ is composed of 20 elements, 10 of which have order 2: ($e$ $\times$ product of 2 two-cycles) and ($(1, 2)$ $\times$ product of 2 two-cycles), and the same pairing but with the 5-cycles of $D_5$, which have orders 5 and 10 depending on whether they are paired with $e$ or $(1, 2)$ ($e$ denotes the identity).

I'm not sure how to proceed from here. My gut tells me the homomorphism should respect the order of the generator $a$ in order to preserve the group structure but that would seem to imply that the only possible homomorphism is the trivial one, which I don't think is true.

Any insight (and perhaps clarification about where I'm going wrong) would be appreciated.

Answer 1

$C_{12}$ is cyclic. So any quotient group (isomorphic to the image) will be cyclic. $C_2\times D_5$ has no cyclic subgroups of order $4$. Therefore the image is either of order $1$ or order $2$.

The map is defined by where to send a generator of $C_{12}$. So either send it to $e$, or send it to one of the elements in $C_2\times D_5$ of order $2$. As you counted, there are $10$ such elements. So there are $11$ such homomorphisms, identified by the elements of order $0$ or order $1$.

Answer 2

Why can't you send $\{a^0, a^2, a^4, a^6, a^{8}, a^{10}\}$ to the kernel and throw everything else to $(1,2) \times \mathrm e$?

Note that an order $12$ element can be sent to an element of order $1$, $2$, $3$, $4$, $6$, or $12$. Consider the map $\phi:\Bbb{Z}/12\Bbb{Z} \rightarrow \Bbb{Z}/4\Bbb{Z}$ where $1_{12} \mapsto 1_{4}$ (and everything else is mapped following the generators: $2_{12} = 1_{12} + 1_{12} \mapsto 1_{4} + 1_{4} = 2_{4}$, and so on).

Answer 3

Since $C_{12}$ is cyclic, the action of $\phi$ is determined by $\phi(e)$ (it seems you're using $e$ for the group identity). If $K=\ker\phi$, you need that $\phi(e)$ has order $|C_{12}/K|$ and any such element defines a homomorphism with that kernel (justify it).

By the homomorphism theorem, we need that, if $K=\ker\phi$, $|K|$ divides $12$ and $12/|K|$ divides $2\cdot10=20$.

In particular, $|K|$ must be a multiple of $3$, so you have just a few choices: $|K|=3$, $|K|=6$ or $|K|=12$.

The last case is trivial. If $|K|=6$, then there are as many homomorphisms as elements of order $2$ in $C_2\times D_5$.

For $|K|=3$, you should find elements of order $4$ in $C_2\times D_5$, of which there's none.