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Problem: Let $f : G → H$ be an isomorphism. Prove that if $a \in G$ has infinite order, then so does $f(a)$, and if $a$ has finite order $n$, then so does $f(a)$. Conclude that if $G$ has an element of some order $n$ and $H$ does not, then $G \not\cong H$.

My proof: If $a$ has finite order $n$, then $a^n=e \Rightarrow f(a^n) = f(a)^n = e$. So, $f(a)$ has order $n$.

How can I prove $a$ had infinite order $\Rightarrow$ $f(a)$ has infinite order?

Minh
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    Have you considered the contrapositive statement? – Shaun Jan 04 '19 at 14:06
  • I know the contrapositive statement, but I couldn't describe it on my proof. – Minh Jan 04 '19 at 14:08
  • The contrapositive statement give $f(a).f(a) \dots \neq e$ – Minh Jan 04 '19 at 14:09
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    Since $f$ is an isomorphism, can't you use $f^{-1}$ combined with the thing you already showed? That should give you the contrapositive unless I glossed over something. –  Jan 04 '19 at 14:10
  • Just because $(f(a))^{n}=e$ does not mean that the order of $f(a)$ is n. You need to further prove that $n$ is the least positive integer which does the above. – ash Jan 04 '19 at 14:11
  • Suppose $a$ has infinite order and $f(a)$ has finite order $n$. We have $f(a)^n = e$ $\Rightarrow$ $f(a^n) = e$ $\Rightarrow$ $a^n = e$, contract with hypothesis. Is that right? – Minh Jan 04 '19 at 14:13
  • If $f$ is an isomorphism, then $f^{-1}$ is also an isomorphism. – Asaf Karagila Jan 04 '19 at 14:16
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    @AshishK Are we need to prove $n$ is the least positive integer with $f(a)^n = e$? – Minh Jan 04 '19 at 14:21
  • @Minh yes consider if $m$ is an positive integer with $m<n$ such that $(f(x))^{m}=e$, can you obtain a contradiction here? – ash Jan 04 '19 at 14:28

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Your proof works in the other direction as well: $$ e = f(a)^n = f(a^n) \overset{f \text{ isomorphism}}{\implies} e = f^{-1}(e) = a^n $$

Jakob E.
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