$G$ Abelian with $H \leq G$. Show $H$ has prime index iff $H$ is kernel of some onto $\phi:G \rightarrow \Bbb{Z}/p\Bbb{Z}$

Question

Let $G$ be an abelian group. Suppose $H$ is a subgroup. Then prove $H$ has index $p$ for some prime if and only if $H:=$ ker$\phi$ where $\phi$ is from $G$ onto $\Bbb{Z} / p \Bbb{Z}$.

For the backwards direction, supposing $H$ is the kernel of some onto group homomorphism

$$\phi:G \rightarrow \Bbb{Z} / p \Bbb{Z}$$

By the first isomorphism theorem we have that $G / H \cong$ im$\phi$ =$\Bbb{Z}/p \Bbb{Z}$ thus $G/H$ has order $p$ thus $H$ has index $p$ in $G$.

For the forwards direction, supposing $H$ has prime index we know that $G/H$ has order $p$. How do I show $H$ Is the kernel of some onto group homomorphism from $G$ onto $\Bbb{Z} / p \Bbb{Z}$? Also is my backwards direction ok?

Answer 1

$H$ is the kernel of the quotient map $\pi: G\to G/H$. Now $G/H$ is of order $p$ and thus it is isomorphic to $\mathbb{Z}/p\mathbb{Z}$. The isomorphism arises from any nontrivial $\mathbb{Z}\to G/H$ map (which is automatically onto because of the prime order) through the first isomorphism theorem. And so if $f:G/H\to \mathbb{Z}/p\mathbb{Z}$ is an isomorphism then $f\circ\pi$ is the map you are looking for.

Answer 2

You need to use the fact that there is only one group with $p$ elements. The reason is that the order of every element $x$ in a finite group must divide the order of the group. Hence if $x$ is non-trivial it must be a generator, and so groups with $p$-elements must be cyclic and isomorphic to $\mathbb{Z}/p\mathbb{Z}$ (say by sending $x$ to $1$ and $x^n$ to $n$ for all $1\leq n \leq p$).

Your proof for the other direction is correct.

Answer 3

Your backwards direction is ok. $G$ does not need to be abelian.

For the forward direction: let $H$ be a normal subgroup of prime index $p$ of a group $G.$ Then, $G/H$ is a group of order $p$ hence isomorphic to $\Bbb Z/p\Bbb Z.$ Let $\varphi:G/H\to\Bbb Z/p\Bbb Z$ be such an isomorphism, and $\pi:G\to G/H$ be the canonical projection. Then, the morphism $\varphi\circ\pi:G\to\Bbb Z/p\Bbb Z$ is onto and its kernel is $H.$