If $\phi:G\to H$ is an isomorphism, prove that $o(x)=o(\phi(x))$ for all $x\in G$

Asked Oct 31 '18 at 22:19

Active Oct 31 '18 at 23:00

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My current thought: $\phi(g^n)=\phi(g)^n$. Thus,$\phi(g^n)=\phi(e)=\phi(g)^n$. However, I don't know how to proceed from here. Also, does my previous steps work when the order of $g$ is infinite?

edited Oct 31 '18 at 23:00

Good Morning Captain

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asked Oct 31 '18 at 22:19

david D

If $G$ has infinite order and $g\in G$ then what are the possible orders of $g$? – Oct 31 '18 at 22:39
Use $\LaTeX$ commands in order to display your equations better. – Doesbaddel Oct 31 '18 at 22:44
@crispypizza must also be infinite, right? – david D Oct 31 '18 at 22:52
While this may seem like the case, it is not always true. Take a look at the quotient group $\mathbb Q/\mathbb Z$. – Oct 31 '18 at 23:01

If $\phi:G\to H$ is an isomorphism, prove that $o(x)=o(\phi(x))$ for all $x\in G$

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