If $G$ is a cyclic group of order $n$ and $p|n$, prove that there exists a homomorphism of $G$ onto a cyclic group of order $p$. Find its Kernel.

My try:

Since $G$ is cyclic and $p|n$ there is a unique subgroup of order $p$ in $G$. So let $G=\langle a\rangle$ and $|G|=n$ and $H=\langle a^m\rangle$ for some $m$ and $|H|=p$. We have to prove that $f:G\to H$ is a homomorphism. For that we need to show that it is $one-one$,$onto$ and $operation-preserving$. How to start? Any hints??

  • $\begingroup$ Try mapping a generator of $G$ to a generator of a cyclic group of order $p$. $\endgroup$ – Lord Shark the Unknown Oct 12 '17 at 21:35
  • $\begingroup$ @LordSharktheUnknown since $H \subseteq G \implies f:G \to G$ but how to show that the function is $1-1$. $\endgroup$ – Vincent Oct 12 '17 at 21:49
  • $\begingroup$ A homomorphism is in general not one-to-one. Choose $m$ such that $n = mp$. $\endgroup$ – amsmath Oct 12 '17 at 21:52
  • $\begingroup$ @amsmath so is it enough to prove that it is operation preserving? $\endgroup$ – Vincent Oct 12 '17 at 21:56
  • $\begingroup$ @VKSingh Yes. But define it first. ;-) $\endgroup$ – amsmath Oct 12 '17 at 22:01

Set $m=n/p$ ($m$ is integer by assumption). Consider the map $$ \varphi\colon G\to G \qquad \varphi(x)=x^m $$ Show

  1. $\varphi$ is a homomorphism
  2. if $g$ is a generator of $G$, then $\varphi(g)\ne1$
  3. $\varphi(G)=\langle g^m\rangle$ is cyclic

Can you finish?

  • $\begingroup$ 1. Let $x,y \in G$ then $\varphi(xy)=(xy)^m=x^my^m=\varphi(x)\varphi(y)$ 2. since $G$ is a cyclic group and $g$ is a generator $g^{n+1}=g$, the $\varphi(g)=\varphi(g^{n+1})=g^{mn+m}=g^m$ 3. every subgroup of a cyclic group is cyclic $\endgroup$ – Vincent Oct 12 '17 at 23:39
  • 1
    $\begingroup$ @VKSingh $g^m\ne1$ because $g^k=1$ implies $n\mid k$. For 3 you don't even need that: since $G$ is cyclic with generator $g$, $\varphi(G)$ is cyclic with generator $\varphi(g)$. On the other hand, $(\varphi(g))^p$ equals… $\endgroup$ – egreg Oct 13 '17 at 6:18

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