# Thread: Prove a group H is isomorphic to Z.

1. ## Prove a group H is isomorphic to Z.

Let H be a group containing an element h with the following property: given a group G and an element g in G, there is a unique homomorphism from H to G mapping h to g. Prove that H is isomorphic to Z (the set of integers).

I know (or I think) that $\displaystyle \psi :H \rightarrow Z$ such that $\displaystyle \psi (h)=1$ defines an isomorphism from H to Z. It's a homomorphism and it is surjective since 1 is a generator of Z. But I'm having a tough time proving that it is injective. Any hints or advice? Thanks.

2. Originally Posted by JD-Styles
Let H be a group containing an element h with the following property: given a group G and an element g in G, there is a unique homomorphism from H to G mapping h to g. Prove that H is isomorphic to Z (the set of integers).

I know (or I think) that $\displaystyle \psi :H \rightarrow Z$ such that $\displaystyle \psi (h)=1$ defines an isomorphism from H to Z. It's a homomorphism and it is surjective since 1 is a generator of Z. But I'm having a tough time proving that it is injective. Any hints or advice? Thanks.
Instead of trying to find an isomorphism you can show that $\displaystyle \mathbb{Z}$ satisfies this property, and also that because of the conditions imposed on $\displaystyle H$ mean it must in fact be $\displaystyle \mathbb{Z}$ (these conditions can be found by remembering that $\displaystyle H$ must map onto $\displaystyle \mathbb{Z}$, the only infinite, one-generated group (up to isomorphism)).

Are you familiar with Free Groups?

3. I'm not familiar with free groups. But I'll give your suggestion a shot. Thanks.

4. as Swlabr mentioned, that's just the universal property of a free group on one generator. but it doesn't mean that the proof is hard! :

consider the maps $\displaystyle H \overset{\psi} \longrightarrow \mathbb{Z} \overset{\phi} \longrightarrow H,$ where $\displaystyle \psi$ is the homomorphism sending $\displaystyle h$ to $\displaystyle 1$ (your map) and $\displaystyle \phi$ is defined by $\displaystyle \phi(n)=h^n,$ for all $\displaystyle n \in \mathbb{Z}.$ note

that $\displaystyle \phi$ is one-to-one. (why?) now consider the maps $\displaystyle \phi \psi : H \longrightarrow H$ and $\displaystyle \text{id}: H \longrightarrow H,$ the identity map. we have $\displaystyle \phi \psi(h)=\phi(1)=h=\text{id}(h).$

thus by the "uniqueness" property: $\displaystyle \phi \psi = \text{id}.$ therefore for any $\displaystyle x \in H$ we have $\displaystyle \phi(\psi(x))=x,$ i.e. $\displaystyle \phi$ is onto and the proof is complete.

5. Wow. Thanks. It's so simple yet I just kept stumbling on it. Thanks a lot.

6. Originally Posted by NonCommAlg
as Swlabr mentioned, that's just the universal property of a free group on one generator. but it doesn't mean that the proof is hard! :

consider the maps $\displaystyle H \overset{\psi} \longrightarrow \mathbb{Z} \overset{\phi} \longrightarrow H,$ where $\displaystyle \psi$ is the homomorphism sending $\displaystyle h$ to $\displaystyle 1$ (your map) and $\displaystyle \phi$ is defined by $\displaystyle \phi(n)=h^n,$ for all $\displaystyle n \in \mathbb{Z}.$ note

that $\displaystyle \phi$ is one-to-one. (why?) now consider the maps $\displaystyle \phi \psi : H \longrightarrow H$ and $\displaystyle \text{id}: H \longrightarrow H,$ the identity map. we have $\displaystyle \phi \psi(h)=\phi(1)=h=\text{id}(h).$

thus by the "uniqueness" property: $\displaystyle \phi \psi = \text{id}.$ therefore for any $\displaystyle x \in H$ we have $\displaystyle \phi(\psi(x))=x,$ i.e. $\displaystyle \phi$ is onto and the proof is complete.
That's a very neat proof. I like it.