0
$\begingroup$

The question from number 2 on page two of this document:

If $\phi: G \to H$ is an isomorphism, prove that $|\phi(x)| = |x|$ for all $x \in G$. Deduce that any two isomorphic groups have the same number of elements of order $n$ for each $n \in \Bbb Z^+$. Is the result true if $\phi$ is only assume to be a homomorphism?


The beginning of the solution states:

Since $\phi: G \to H$ is a homomorphism, $\phi (x^n) = \phi(x)^n$ for all $n \in \Bbb Z^+$. (I assume this is true because we can alternatively write it as $\phi (x*x*x...x) = \phi(x) * \phi(x) * \phi(x) * ... * \phi(x)$ where the right hand side is computed in $H$ and the left hand side is computed in $G$)

When $n = 0$, we have $\phi(1_G) = 1_H)$.

If $|x| = m$ is finite, then $x^m = 1_G$ and so $\phi(x)^m = 1_H$, which shows $|\phi (x)| \leq |x|$ (since isomorphism implies bijection).

I see how if $x$ has finite order, then $x^m = 1_G$, but why can we deduce that $\phi(x)^m = 1_H$ if $x^m = 1_G$?

Thanks for the help!

$\endgroup$
0
2
$\begingroup$

An isomorphism sends only the identity to the identity, so $$\phi(x)^m = \phi(x^m) = \phi(1_G) = 1_H.$$

By the way, not only do you have $|\phi(x)| \leq |x|$ but you actually have $|\phi(x)| \mid |x|$, i.e. the order of $\phi(x)$ divides the order of $x$. (This requires just a bit of work.)

$\endgroup$
1
  • $\begingroup$ I saw your comment earlier so you get the check! The textbook I'm reading doesn't make it clear that isomorphism sends the identity to the identity. Thanks for the help. $\endgroup$
    – Raoul Duke
    Feb 7 '20 at 4:26
1
$\begingroup$

Suppose $\phi(x)^n=1_H$, where $n$ is the order of $\phi(x)$. Then $\phi(x^n)=1_H$. Since $\phi$ is injective, we have $x^n=1_G$. Hence $|x|\leq |\phi(x)|$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.