0
$\begingroup$

Let $H$ be a group and suppose that $ f: D_{10} \rightarrow H $ is a homomorphism. How do I describe and justify all the possible images of $f$.

$D_{10} = ({1, \sigma, \sigma^2, \sigma^3, \sigma^4, \tau, \sigma\tau, \sigma^2\tau, \sigma^3\tau, \sigma^4\tau}) $ where $\sigma = (12345)$ and $\tau = (13)(45)$. I've been stuck on this for a while. I was given a hint to use the 1st Isomorphism Theory but that fell through. Please help! I have found that normal subgroups of $ D_{10}$ are $\left \{ e \right \}, \left \{ D_{10} \right \} and \left \langle \sigma \right \rangle $ but I've hit a wall.

Added: I've learnt that the First Isomorphism Theorem tells us that if $ f: G \rightarrow H$ is a homomorphism then $G/Ker(f) \cong Im(f)$. Thus in the current example, as $ f: D_{10} \rightarrow H$ has been defined as a homomorphism then $D_{10}/Ker(f) \cong Im(f)$. I've also been informed that the Kernels are the normal subgroups of $D_{10}$ which I found were $\left \{ e \right \}, \left \{ D_{10} \right \} and \left \langle \sigma \right \rangle $. So I need to find $D_{10}/\left \{ e \right \}, D_{10}/\left \{ D_{10} \right \}$ and $D_{10}/\left \langle \sigma \right \rangle $. But I'm stuck as to how.

$\endgroup$
2
$\begingroup$

Hint: Recall that the First Isomorphism Theorem tells you that if $f:G\to H$ is a homomorphism, then $G/\ker f\cong \text{Im } f$. I assume $H$ can be any arbitrary group, so we just need to think about what possible kernels could be. But remember that these are just the normal subgroups of $G$ (they are kernels of, for instance, the natural projections onto the induced quotient group). Find the normal subgroups of $G$ and mod out by them to find all possible images.

$\endgroup$
  • $\begingroup$ Apologies, I've added the normal subgroups of $D_{10}$ that I've found. $\endgroup$ – Geometry Dec 4 '15 at 22:44
  • $\begingroup$ You have all of them, so that's good! Now you can easily tell what the images are. For the case of $\{e\}$ and $\{D_{10}\}$, you should be able to find the trivial quotients. For $D_{10}/\langle \sigma\rangle$, what is the order of this quotient? What is the only group of that order? $\endgroup$ – J.G Dec 4 '15 at 22:49
  • $\begingroup$ I'd like to thank you for guiding me. I've only just started MathStack so it's lovely that people are willing to aid others without any direct requirement for reciprocation. I'm really struggling with the concept of an image. Would you be able to describe exactly what it is? $\endgroup$ – Geometry Dec 4 '15 at 23:11
  • $\begingroup$ It's my pleasure! So you can think of the image of a function to be all of the elements that are mapped to. So if we had a map that took every integer and doubled it, the image would be all of the even integers. If we had a map that took every element of a group and mapped to the identity of another group, the image would just be the identity (this is the trivial homomorphism). Hope this helps! $\endgroup$ – J.G Dec 4 '15 at 23:34
  • $\begingroup$ In this case the map is from $ D_{10} \rightarrow H $, could you give me an example of an image? I think I'm having trouble getting over the fact that it's just mapping to $H$ with no other description. $\endgroup$ – Geometry Dec 4 '15 at 23:39
0
$\begingroup$

I think you must find all the normal subgroups of $D_{10}$ then use the fact that every normal subgroup is the kernel of a group homomorphism of $D_{10}$. i.e. $ker f = N$, where $N \unlhd D_{10}$. I hope that answer your question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.