# What is wrong with my proof of group order?

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.

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 $$k such that $$(\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 $$r is order of $$r$$

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

• 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.
– anon
Commented Mar 25, 2022 at 15:11
• Would be good: mention how to define $m$ and $r$. I guess $m, r ∈ ℕ$, $m \geq 1$, $k > r \geq 0$. Commented Mar 25, 2022 at 15:12
• 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. Commented Mar 25, 2022 at 15:13
• @runway44 it used the isomorphism when he claims $\phi (k) = e_h \implies a^k = e_g$. Commented Mar 25, 2022 at 15:13
• @runway44 makes an excellent point worth remembering. Be wary if you find yourself not using the entire hypothesis to prove a claim! Commented Mar 25, 2022 at 15:13

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 $$b$$
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 $$n = k$$.