Prove a group H is isomorphic to Z.

Printable View

September 22nd 2009, 08:34 PM
JD-Styles

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 $\psi :H \rightarrow Z$ such that $\psi (h)=1$ 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.
September 23rd 2009, 12:56 AM
Swlabr

Quote:

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 $\psi :H \rightarrow Z$ such that $\psi (h)=1$ 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 $\mathbb{Z}$ satisfies this property, and also that because of the conditions imposed on $H$ mean it must in fact be $\mathbb{Z}$ (these conditions can be found by remembering that $H$ must map onto $\mathbb{Z}$ , the only infinite, one-generated group (up to isomorphism)).

Are you familiar with Free Groups?
September 23rd 2009, 06:09 AM
JD-Styles

I'm not familiar with free groups. But I'll give your suggestion a shot. Thanks.
September 23rd 2009, 12:40 PM
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! (Shake):

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

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

thus by the "uniqueness" property: $\phi \psi = \text{id}.$ therefore for any $x \in H$ we have $\phi(\psi(x))=x,$ i.e. $\phi$ is onto and the proof is complete.
September 23rd 2009, 05:48 PM
JD-Styles

Wow. Thanks. It's so simple yet I just kept stumbling on it. Thanks a lot.
September 23rd 2009, 11:08 PM
Swlabr

Quote:

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! (Shake):

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

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

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

That's a very neat proof. I like it.