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I want to know the possible images of group homomorphisms that map $D_{10} \rightarrow G$, where $G$ is some arbitrary group.

Things I know: $\frac{G}{ker(f)} \cong Im(f)$; $D_{10}$ normal subgroups are $\{e\}$, $<\sigma>$ and $D_{10}$.

I do not know how to connect all this information, so please help!

Mathmo123
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sara
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    A few suggestions. Use TeX properly, your question is hard to read. Also: define your notation, your question makes no sense otherwise. I have no idea what $sigma$ is. Also, perhaps I could guess what D10 is, but I might be wrong and so I am not inclined to answer. – Lee Mosher Dec 03 '15 at 17:21

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You're basically there! Suppose $\theta:D_{10}\to G$ is a group homomorphism. Since we're only interested in the image of $\theta$, we might as well assume that $\theta$ is surjective, so that $G=\mathrm{Im}(\theta) $. Then by the first isomorphism theorem, $$G\cong D_{10}/\ker\theta$$which isn't quite what you stated.

Hence, if $G$ is the image of a such a homomorphism, then we must have $G\cong D_{10}/N$ for some normal subgroup $N$.

The converse is also true: if $N$ is a normal subgroup, then the map$$D_{10}\to D_{10}/N\\g\mapsto gN$$is a homomorphism.

Hence the possible images are exactly $D_{10}/N$ where $N$ is normal in $D_{10}$.

Mathmo123
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  • Thanks a lot! you are very helpful! please just help me understand how to find the quotient groups - this part really confuses me! – sara Dec 03 '15 at 17:33
  • Do you know the definition of a quotient group? You've calculated all the normal subgroups, and each normal subgroup gives a unique quotient group. – Mathmo123 Dec 03 '15 at 17:34
  • for example - one of the normal subgroups is {e}, therefore D10/{e} = a{e} for all a in D10, which is D10? – sara Dec 03 '15 at 17:39
  • however D10 is also a normal subgroup, so aD10 for all a in D10 is what exactly? this is where I get confused – sara Dec 03 '15 at 17:40
  • If $g\in D_{10}$, then $gD_{10} = D_{10}$, so there is only one left coset. So the quotient group is trivial. – Mathmo123 Dec 03 '15 at 17:41
  • so for {e} the number of cosets is 10, for D10 the number of left cosets is 1! but I didn't think it matter about the number of left cosets - why is that relevents? also for $$ the number of left cosets is 2 – sara Dec 03 '15 at 17:47
  • The underlying set of the quotient group is just the set of left cosets. Knowing how many of them there are can often tell you what the quotient group looks like. For example, since there is only one group of order 2 (up to isomorphism), $D_{10}/\langle\sigma\rangle$ is just the 2-element cyclic group – Mathmo123 Dec 03 '15 at 17:49
  • ok, so that means that D10/$$ = {e, t}, {e, st}, {e,s^2t}, {e, s^3t}, {e, s^4t} ?? – sara Dec 03 '15 at 17:56
  • and D10/D10 only has 1 left coset so what does that mean! I feel like I am almost there! thank you so much – sara Dec 03 '15 at 17:58
  • No. It is ${\langle \sigma \rangle, t\langle\sigma\rangle}$ – Mathmo123 Dec 03 '15 at 17:58
  • ok, please don't leave - I need your help! I understand why that looks like that - so far we have D10, D10 , {, t}? – sara Dec 03 '15 at 18:03
  • You might want to look again at the definition of the quotient group. That should help you more than me telling you the answers. – Mathmo123 Dec 03 '15 at 18:07
  • I know the definition, but I can't get my head round what it actually looks like - I have no examples to look at – sara Dec 03 '15 at 18:09
  • Take a look at http://math.stackexchange.com/questions/69050/why-the-term-and-the-concept-of-quotient-group and https://gowers.wordpress.com/2011/11/20/normal-subgroups-and-quotient-groups/ – Mathmo123 Dec 03 '15 at 18:17