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Let p be a prime and let $\displaystyle \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 $\displaystyle f: \mathbb {Z} (p^ \infty ) \rightarrow G $ is a surjective group homomorphism and G is not the trivial group $\displaystyle \{ 1_G \} $, then G is isomorphic to $\displaystyle \mathbb {Z} (p^ \infty ) $.
this is one of interesting properties of $\displaystyle \mathbb{Z}_{p^{\infty}}. $ suppose $\displaystyle \ker f = K.$ then since $\displaystyle f$ is surjective, we have $\displaystyle \frac{\mathbb{Z}_{p^{\infty}}}{K} \simeq G.$ since $\displaystyle G$ is nontrivial, we have $\displaystyle K \neq \mathbb{Z}_{p^{\infty}}.$ so what we need to prove is that for anyQuote:
Originally Posted by tttcomrader
subgroup $\displaystyle I$ of $\displaystyle \mathbb{Z}_{p^{\infty}}$ with $\displaystyle I \neq \mathbb{Z}_{p^{\infty}}$ we have $\displaystyle \frac{\mathbb{Z}_{p^{\infty}}}{I} \simeq \mathbb{Z}_{p^{\infty}}.$ we have proved before that we must have $\displaystyle I=<\hat{\frac{1}{p^k}}>,$ for some $\displaystyle k \geq 0.$ now define $\displaystyle \varphi: \mathbb{Z}_{p^{\infty}} \longrightarrow \mathbb{Z}_{p^{\infty}}$ by $\displaystyle \varphi(\hat{\frac{a}{p^n}})=\hat{\frac{a}{p^{n-k}}}.$ see that $\displaystyle \varphi$ is a
well-defined homomorphism. it's also surjective. finally $\displaystyle \ker \varphi = I,$ which is again obvious because: $\displaystyle \hat{\frac{a}{p^n}} \in \ker \varphi \Longleftrightarrow \frac{a}{p^{n-k}} \in \mathbb{Z} \Longleftrightarrow \frac{a}{p^n} \in \frac{1}{p^k} \mathbb{Z} \Longleftrightarrow \hat{\frac{a}{p^n}} \in <\hat{\frac{1}{p^k}}>=I.$ so we've proved that $\displaystyle \varphi$ is a
surjective homomorphism and its kernel is $\displaystyle I.$ thus by the first isomorphism theorem for groups we must have $\displaystyle \frac{\mathbb{Z}_{p^{\infty}}}{I} \simeq \mathbb{Z}_{p^{\infty}}. \ \Box$
