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

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?

$$H$$ 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.
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 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 $$H.$$