Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Check if there exists an isomorphism for : $$ i) \ G = (\mathbb{Z}_{15}, +_{15}), \ H = (\mathbb{Z}_{15}, +_{15}), \ f(1) = 2 $$ $$ ii) \ G = (\mathbb{Z}_{15}, +_{15}), \ H = (\mathbb{Z}_{15}, +_{15}), \ f(1) = 3 $$ $$ iii) \ G = (\mathbb{Z}_6, +_6), \ H = S_3 $$ $$ iv) \ G = (\mathbb{R}, +), \ H = (\mathbb{Z}, +) $$ where $$ f: G \rightarrow H $$

Problem: I know that I should firstly check conditions for homomorphism and this part I can, but I don't know how I have to check surjection and injection. Could you help me ;) ?

share|cite|improve this question
Which of the examples have you checked the existence of a (non-trivial) homomorphism for? Have you ruled any of the answers out? – Devlin Mallory Aug 16 '13 at 21:05
Do you know a condition equivalent to injectivity regarding the kernel of a homomorphism? Do you know the first isomorphism theorem? It could be helpful. – Stefan Hamcke Aug 16 '13 at 21:06
@DevlinMallory just already check i) - it can't be, because $$ f(14 +_{15} 1) = f(0) = 2 != 0 $$ (must exist e), example ii) is correct, but I have problem with iii), because I don't know how I should start check homomorphism could you give me hint? – Mat Aug 16 '13 at 21:14
Your check for part (i) doesn't make any sense to me. Take a look at my answer below and see if it helps you. – Cameron Buie Aug 16 '13 at 21:16
I would double check your reasoning for i) and ii); I'm not sure what you mean by $f(0)=2!=0$, but an isomorphism should exist in i) and not ii). Otherwise, Cameron Buie gives a great answer below – Devlin Mallory Aug 16 '13 at 21:16
up vote 1 down vote accepted

Well, in the first three parts, $1$ generates $G,$ so given any $h\in H,$ there is a unique homomorphism $f:G\to H$ such that $f(1)=h,$ given by $$f(\underset{n\text{ times}}{\underbrace{1+_{15}\cdots+_{15}1}})=\underset{n\text{ times}}{\underbrace{h+_{15}\cdots+_{15}h}}.$$ (This is a nice exercise.) Hence, there's no need to confirm that $f$ is a homomorphism in the first two cases. Among finite sets of the same size, functions are injective if and only if they are surjective, so you need only check one of those in each the first two cases (I recommend injectivity, personally.)

For the third part, think about what you know about $G$ and $H.$ Do you know anything different about them that would make an isomorphism impossible? If not, I recommend that you figure out what the homomorphisms $G\to H$ are, and see if any are injective/surjective (remember, we only need to choose where to send $1,$ and the homomorphism is completely determined).

For the fourth part, think about what you know about $G$ and $H.$ Do you know anything different about them that would make an isomorphism impossible? Specifically, what do you know about the relative sizes of the two sets in question?

share|cite|improve this answer
Ok, I read your answer and I check the homomorphism for i) and ii) your method and it's good solution but I don't know how I should check injectivity, because I've never check before. Could you show me for i) ? – Mat Aug 16 '13 at 21:34
You need to show that if $f(x)=f(y),$ then $x=y.$ Alternately, show that if $x\ne y,$ then $f(x)\ne f(y)$. Before I show you, I note that you mentioned above that $f$ is not a homomorphism in the second case. How did you determine that? – Cameron Buie Aug 16 '13 at 21:36
Yes, again my fault I didn't write number 15, and this was a bug ;) I checked again and i) and ii) is correct. Your method is awesome! – Mat Aug 16 '13 at 21:42
Do you know what the kernel of a homomorphism is? If so, what are some things you know about it? – Cameron Buie Aug 16 '13 at 21:50
I heard about this, but I've never used in a practial. – Mat Aug 16 '13 at 21:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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