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If $f:G\rightarrow H$ is an isomorphism, then $\forall g \in G$, we have that the order of $g$ is the same as the order of $f(g)$

I can understand intuitively why this is true, but what would be the clearest way of proving this? Would you use some contradiction or would you define a bijection from the set generated by $g$ to the set generated by $f(g)$?

Henry Brown
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  • Are you aware that if $g$ is an element of some group $G$ and if it happens that $g^{k}$ is the identity element for some integer $k$ then order of $g$ divides $k$? – ash Nov 15 '20 at 09:02
  • Yes, I do know that result. Could you use that to form a contradiction? – Henry Brown Nov 15 '20 at 09:04

2 Answers2

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Let $g\in G$ with order $n$. Then $e_H=f(e_G) = f(g^n) = f(g)^n$. So the order of $f(g)$ is a divisor of $n$.

Let $d$ be the order of $f(g)$. Then $d$ divides $n$. Thus $e_H = f(g)^d = f(g^d)$ and hence $f(e_G) =e_H= f(g^d)$. But $f$ is an isomorphism (one-to-one is required here) and so $e_G=g^d$. By hypothesis ($d$ divides $n$), $d=n$.

Wuestenfux
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    This reasoning doesn't cover the general case of a given element of potentially infinite order. – ΑΘΩ Nov 15 '20 at 09:08
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    How would you deal with the case where G and H are not finite and you have an element g which has infinite order? Edit: ΑΘΩ got there just before me! – Henry Brown Nov 15 '20 at 09:09
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Let $f \in \mathrm{Hom}_{\mathbf{Gr}}(G, G')$ be an injective morphism between the groups $G$ and $G'$. It is a fundamental property of morphisms (not necessarily injective) that $f\left[\langle X \rangle\right]=\langle f[X] \rangle$ for any subset $X \subseteq G$, where $\langle X \rangle$ denotes the subgroup of $G$ generated by $X$.

Let us fix $x \in G$. By definition, the order of $x$ is $\left|\langle x \rangle\right|$, the order of the monogenous subgroup generated by $x$. Taking into account the general property mentioned above, we have $\langle f(x) \rangle=f\left[\langle x \rangle\right]$. As $f$ is injective, it follows by elementary cardinal theory that $|\langle f(x) \rangle|=\left|f\left[\langle x \rangle\right]\right|=|\langle x \rangle |$ (in general $|f[T]|=|T|$ for any subset $T$ of the domain of definition of the injection $f$). This means nothing else than the relation you seek to prove: the order of $x$ and the order of its image $f(x)$ coincide.

ΑΘΩ
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  • Exactly what I'm looking for – Henry Brown Nov 15 '20 at 09:27
  • Where is the surjectivity of the isomorphism used here? –  Nov 15 '20 at 16:35
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    @hal Nowhere, because it isn't required. One of my principles is that of trying to deduce a maximum of claims with only a minimum of hypotheses, and it does apply to this particular situation. – ΑΘΩ Nov 16 '20 at 00:06