Let's say $f$ is an isomorphism $f:G \rightarrow G'$, where $G$ and $G'$ are multiplicative groups.
Then for $x\in G$, if $f(x) = x' \in G'$. Do we have always have $\text{ord}(x) = \text{ord}(x')$? Why?
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6Just a side note: $f(x\in G)$ is terrible notation, write something like "$f(x)=x'$, where $x\in G$" instead. – user2520938 Jul 10 '17 at 17:41
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By $o(x)$, do you mean the order of $x$? – SvanN Jul 10 '17 at 17:42
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@user2520938 Ok, thank you ! – Littletry Jul 10 '17 at 17:58
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Put $y = f(x)$. If $o(x) = n$, then we have $y^n = f(x)^n = f(x^n) = 1$, so that $o(y)\le n$. If $m < n$ and $1 = y^m = f(x)^m = f(x^m)$, then applying $f^{-1}$ to both sides of this equation and noting that isomorphisms send $1\mapsto 1$, we have $1 = x^m$, absurd.
Alex Ortiz
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Nice one ! Thank you ! Just : $f(x)^n = f(x^n)$ is true for every isomorphisme even with non-multiplicative groups ? – Littletry Jul 10 '17 at 18:01
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3@Littletry Note that every isomorphism is, in particular, a homomorphism, so that $f(ab) = f(a)f(b)$, and by induction $f(x^n) = f(x)^n$. For groups written additively, this equation becomes $f(a+b) = f(a)+f(b)$, and by induction $f(nx) = nf(x)$ where $nx$ means $x + x + \dotsb + x$ ($n$ times). – Alex Ortiz Jul 10 '17 at 18:03
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Why is that absurd? It seems like you just found a smaller solution to the order, implying that |x|=m for some m < n. I thought you wanted to show m=n by getting a contradiction?? – John D Feb 16 '24 at 02:00
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@JohnD, by definition of $n$, there can be no smaller $m$ such that $x^m=1$. Another way to say it is if $x^m=1$, then by definition of $n$ as the order of $x$, $m\ge n$. This is a contradiction because we started assuming $m<n$. – Alex Ortiz Feb 16 '24 at 03:40
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"Isomorphism" means that the elements of $G$ obtain new names taken from the set $G'$, but everything operationwise stays the same. It is therefore innate in the very essence of "isomorphism" that the claim you are told to prove is true. Any "formal proof" would make the claim less believable.
Christian Blatter
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More generally, if $f:G \rightarrow G'$ is a group homomorphism, then $o(f(x))$ divides $o(x)$ simply because $o(x)=m$ implies $e'=f(e)=f(x^m)=f(x)^m$. In particular, $o(f(x)) \le o(x)$.
When $f$ is an isomorphism with inverse $g$, then we get $o(x) = o(gf(x))\le o(f(x)) \le o(x)$ and so $o(f(x)) = o(x)$.
lhf
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You show that |f(x)| ≤ |x| twice, you probably meant to say |f(x)| ≥ |x| at some point to conclude they are equal but your proof as it stands is wrong. – John D Feb 16 '24 at 01:40
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@JohnD, I don't think so. I use that $o(h(z)) \le o(z)$ twice: once for $h=f$ and $z=x$ and once for $h=g$ and $z=f(x)$. – lhf Feb 16 '24 at 02:34
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Yes. Isomorphisms preserve order. In fact, any homomorphism $\phi$ will take an element $g$ of order $n$ to an element of order dividing $n$, by the homomorphism property. Now since an isomorphism has an inverse which is also a homomorphism, the claim follows. For we get that $n|o(\phi(g))$.
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The claim holds also for infinite $\langle g\rangle$. In fact, by the injectivity, $\langle g\rangle\cong f(\langle g\rangle)$; moreover, $f(\langle g\rangle)\le\langle f(g)\rangle$ generally holds for any homomorphism. If $\langle g\rangle$ is infinite, then $\langle g\rangle\cong f(\langle g\rangle) \cong \Bbb Z$, and hence $\langle f(g)\rangle\cong \Bbb Z$. Therefore, $|\langle g\rangle|=|\langle f(g)\rangle|$.