Find all $n$ for which exist homomorphism onto $(\mathbb {Z} _n, +)$. (It means find surjective homomorphism.)

(a) $(\mathbb {D}_3,\circ)$ where $\mathbb {D}_3$ is the symmetry group of an equilateral triangle.

(b) $(\mathbb {R}, +)$.

I know that an homomorphism $f$ into $\mathbb{Z}_n$ is surjective if and only if $\operatorname{Im}(f)=\mathbb{Z}_n$. And a map $f$ is an homomoprhism from the group $(G,\circ )$ into the group $(H,*)$ if $\forall a,b ∈ G, f(a \circ b)=f(a)*f(b)$.

But I do not have any idea how to find all $n$ and how to start.

  • $\begingroup$ Please ask one question at a time. $\endgroup$ – Shaun Dec 30 '17 at 16:48
  • $\begingroup$ Use $\circ$ for $\circ$. $\endgroup$ – Shaun Dec 30 '17 at 16:49
  • 1
    $\begingroup$ Hint: if $f$ is onto, then $\bar 1$ belongs to image of $f$. But $\bar 1$ has order ... $\endgroup$ – Sigur Dec 30 '17 at 16:58
  • $\begingroup$ @Waney:I'm not getting your question,what is the domain of surjection?Am i right?? $\endgroup$ – P.Styles Dec 30 '17 at 17:11

For (a): Note that, if $\sigma{}$ is the rotaion by 120 degrees and $\tau{}$ is the reflection, and $\phi{}:D_3\to{}\mathbb{Z}_n$ is a homomorphism, then


So $\phi(\sigma)=\phi(\sigma)^{-1}$, as $\mathbb{Z}_n$ is abelian, and thus the order of $\phi(\sigma)$ is 1 or 2.

$D_3$ has 6 elements, so necessairly $n\le{}6$. Also, we know that $n=|Im(\phi{})|$ must divide $|D_3|=6$ from the first isomorphism theorem. So we have the options $n=1,2,3,6$. 6 doesn't work because the groups aren't isomorphic. 3 doesn't work because, as we said, the order of $\phi(\sigma)$ must divide 2, it can't be two because no element of $\mathbb{Z}_3$ has order 2 and it also can't be 1 (why?). So we have $n=1$ or $n=2$, and it is easy to find the surjective homomorphisms for them.

For (b): Note that if $\phi:\mathbb{R}\to\mathbb{Z}_n$ is a homomorphism, then $\phi(nx)=\phi(x)+...\phi(x)=0$ for all $x\in\mathbb{R}$, and since $nx$ can be every value, $\phi$ must be the trivial homomorphism, so $n=1$ is the only solution.


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