Let $\phi: G\rightarrow H$ be an isomorphism of groups and let $a \in G$ be of order $n$. Show that the order of $\phi(a)$ is also $n$.

I was given this problem a week ago during a quiz and my following answer has been graded as "partially" correct lately. Despite checking my proof for a decent time, I could not figure out why my proof cannot be considered as "fully" correct so I have decided to ask it in here with my proof.


$a\in G$ is order of $n$ $\Rightarrow$ $a^n=e_{g}$

$a^n=e_g \Rightarrow \phi(a^n)=e_{h}$

$\phi(a^n)=(\phi(a))^n \Rightarrow (\phi(a))^n=e_h$

Now assume that there exist a $k<n$ such that $(\phi(a))^k=e_h$ and $k$ is order of $\phi(a)$

$(\phi(a))^k=e_h \Rightarrow (\phi(a))^n=(\phi(a))^{mk+r}=(\phi(a))^r=e_h$

$(\phi(a))^r=\phi(a^r)=e_h \Rightarrow a^r \in \ker\phi$

$a^r \in \ker\phi \Rightarrow a^r=e_g$

$a^r=e_g$ and $r<k \Rightarrow a $ is order of $r$

This is obviously a contradiction hence order of $\phi(a)$ must be $n$

  • 2
    $\begingroup$ Notice your "proof" did not use the fact $\phi$ is an isomorphism, only the fact it is a homomorphism. What can go wrong? If $\phi$ is not an isomorphism, then $\phi$ can send $a$ of order $n$ to $\phi(a)$ of order strictly dividing $n$, in which case $r=0$, but then the fact $a^r=e_G$ and $r<k$ does not yield a contradiction. $\endgroup$
    – anon
    Commented Mar 25, 2022 at 15:11
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    $\begingroup$ Would be good: mention how to define $m$ and $r$. I guess $m, r ∈ ℕ$, $m \geq 1$, $k > r \geq 0$. $\endgroup$
    – flukx
    Commented Mar 25, 2022 at 15:12
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    $\begingroup$ You seriously overcomplicated the second part of the proof by introducing the unnecessary $r$, and then forgot the case $r=0$. You could just use $\phi(a^k)=e_h$ to conclude. $\endgroup$ Commented Mar 25, 2022 at 15:13
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    $\begingroup$ @runway44 it used the isomorphism when he claims $\phi (k) = e_h \implies a^k = e_g$. $\endgroup$ Commented Mar 25, 2022 at 15:13
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    $\begingroup$ @runway44 makes an excellent point worth remembering. Be wary if you find yourself not using the entire hypothesis to prove a claim! $\endgroup$
    – Gregory
    Commented Mar 25, 2022 at 15:13

2 Answers 2


The problem is that, if $k$ divides $n$, you end up with $\big(\phi(a)\big)^0=e_h$, which is trivially true, because $b^0=e_h$ for all $b$


In addition to @Evaristos answer here the completion/correction:

Let $k = \operatorname{ord} φ(a)$. Since $e_h = φ(e_G) = φ(a^n) = φ(a)^n$, $n \geq k$. $φ(a^k) = φ(a)^k = e_h$. Since $φ$ is bijective and $φ(e_h) = e_G$, we get $a^k = e_G$. Hence $n | k$ but with $k \leq n$ we have $n = k$.


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