(For any group $H$) Bijection between $H$ and the group consists of all the homomorphisms between $(\mathbb Z, +)$ and $H$.

Question

My question is, for any group $H$, how to prove there exists a bijective function between $Z$ and the group consists of all the homomorphisms which I will call IS later on for convenience

I just started learning abstract algebra and soon I encountered this problem.

Find a group $G$ such that, for any group $H$, there is always a bijection between the set of group homomorphisms $\phi : G \rightarrow H$ and the set of elements of $H$. Note: $G$ is independent of $H$.

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I have known the $G$ is $Z$ and what I need to do is to construct a function and prove it bijective.

Define $i_h(n) = h^n$ and obviously $i_h$ is a homomorphism.

So I can construct a function named $b$ which is from $H$ to $IS$. Thence $\mathrm { b } : \mathrm { H } \rightarrow \mathrm { IS }$

$$ b(h) = i_h $$

And finally the last two tasks.

a) injective

$\forall h _ { 1 } , h _ { 2 } \in H \text { if } b \left( h _ { 1 } \right) = b \left( h _ { 2 } \right) \text { we will have } b \left( h _ { 1 } \right) ( 1 ) = b \left( h _ { 2 } \right) ( 1 ) \text { which will result in the fact } h _ { 1 } = h _ { 2 }$

b) surjective

But I don't know how to prove the surjective part.

Answer 1

For any $\phi\in IS$, we have $h:=\phi(1)\in H$. We will claim that $$b(h) = i_h=\phi$$ as follows:

$$i_h(n)=h^n=(\phi(1))^n=\phi(n*1)=\phi(n)$$ for all $n\in \mathbb{Z}$. Thus $i_h=\phi$. This means that $\phi=b(h)$ is in the image of $b$. So $b$ is surjective.

Answer 2

You have to prove that $IS=\{i_{h}, h\in H\}$, $i_{h}$ defined above. $\mathbb{Z}$ is the free group generated from the element 1, so an homomorphism $\mathbb{Z}\to H$ is defined by the imagine of 1, and in your set of homomorphisms there are all the possible imagines for the element 1.

Answer 3

For surjectivity, let $f \in \text{IS}$. Now take $h=f(1)$, then $b(h)=i_{h}$. Hence surjective.