Isomorphism of a surjective group homomorphism

Quote:

Originally Posted by tttcomrader

Let p be a prime and let $\mathbb {Z} (p^ \infty ) = \{ \hat { \frac {a}{b} } \in \frac { \mathbb {Q} }{ \mathbb {Z} } : a,b \in \mathbb {Z} \ , \ b=p^i \ , \ i \geq 0 \}$ .

Prove that if $f: \mathbb {Z} (p^ \infty ) \rightarrow G$ is a surjective group homomorphism and G is not the trivial group $\{ 1_G \}$ , then G is isomorphic to $\mathbb {Z} (p^ \infty )$ .

this is one of interesting properties of $\mathbb{Z}_{p^{\infty}}.$ suppose $\ker f = K.$ then since $f$ is surjective, we have $\frac{\mathbb{Z}_{p^{\infty}}}{K} \simeq G.$ since $G$ is nontrivial, we have $K \neq \mathbb{Z}_{p^{\infty}}.$ so what we need to prove is that for any

subgroup $I$ of $\mathbb{Z}_{p^{\infty}}$ with $I \neq \mathbb{Z}_{p^{\infty}}$ we have $\frac{\mathbb{Z}_{p^{\infty}}}{I} \simeq \mathbb{Z}_{p^{\infty}}.$ we have proved before that we must have $I=<\hat{\frac{1}{p^k}}>,$ for some $k \geq 0.$ now define $\varphi: \mathbb{Z}_{p^{\infty}} \longrightarrow \mathbb{Z}_{p^{\infty}}$ by $\varphi(\hat{\frac{a}{p^n}})=\hat{\frac{a}{p^{n-k}}}.$ see that $\varphi$ is a

well-defined homomorphism. it's also surjective. finally $\ker \varphi = I,$ which is again obvious because: $\hat{\frac{a}{p^n}} \in \ker \varphi \Longleftrightarrow \frac{a}{p^{n-k}} \in \mathbb{Z} \Longleftrightarrow \frac{a}{p^n} \in \frac{1}{p^k} \mathbb{Z} \Longleftrightarrow \hat{\frac{a}{p^n}} \in <\hat{\frac{1}{p^k}}>=I.$ so we've proved that $\varphi$ is a

surjective homomorphism and its kernel is $I.$ thus by the first isomorphism theorem for groups we must have $\frac{\mathbb{Z}_{p^{\infty}}}{I} \simeq \mathbb{Z}_{p^{\infty}}. \ \Box$