2

My reference for learning abstract algebra says that in an abelian group $G$ the product of two elements (say $a$ with order $n$ and $b$ with order $m$) has order $mn$ if $gcd(n,m) = 1$.

What I don't understand is where the $gcd(m,n)=1$ comes into use in proving this fact? For example, $ a^{mn} = (a^n)^m = (e)^m = e$ (identity element). A similar argument for $b^{mn}$ together with the fact that $G$ is abelian is enough to show that $(ab)^{mn} = e$. From here, how do I conclude for an integer $r < mn$ that $(ab)^r \neq e$

Also, are there any other conclusion that can be drawn if $gcd(n,m) \neq 1$ ?

Thanks for any feedback in advance !

Lgate8
  • 445

1 Answers1

3

If $(ab)^k=1$ then $a^k b^k=1$, because $G$ is abelian, and so $a^k = b^{-k}$.

Now, $a^k = b^{-k}$ implies that $a^k,b^k \in \langle a \rangle \cap \langle b \rangle = 1$, because these subgroups have coprime orders.

Therefore, $k$ is a multiple of both $n$ and $m$, and so is a multiple of $lcm(n,m)=nm$.

If $k$ is the least integer such that $(ab)^k=1$, then we must have $k=nm$.

lhf
  • 216,483