orders of groups

Question

Not really sure how to go about this in general terms, for example: say you have a group $G$ containing $a$ and $b$ such that $a\cdot b=b \cdot a$. Show that if $a$ has order $m$ in and $b$ has order $n$ ($\in \mathbb{N}$), and $\gcd(m,n)=1$, then $a \cdot b$ has order $mn$.

Answer 1

If $(ab)^k=1$, then $a^k = b^{-k}$.

Now $o(a^k)$ divides $o(a)=m$ and $o(b^{-k})$ divides $o(b)=n$.

Therefore, $o(a^k)$ divides $\gcd(m,n)=1$, and so $a^k=1$. This implies that $k$ is a multiple of $m$.

Also, $b^{-k}=1$ implies that $k$ is a multiple of $n$.

Since $\gcd(m,n)=1$, we must have that $k$ is a multiple of $mn$.

Since $(ab)^{mn}=1$, the order of $ab$ is $mn$.

Answer 2

Hint: Note that $$ (ab)^k = a^kb^k $$ for all integers $k$. Use this to show that $(ab)^k = e$ if and only if $k$ is a multiple of the orders of both $a$ and $b$.

Answer 3

We know that $a$ has order $m$ (we denote $o(a) = m$) and $o(b) = n$. We have that $$(ab)^{mn} = \underbrace{ab \cdot ab \cdot \ldots \cdot ab}_{mn \text{ times}}$$ and since $ab = ba$ in the group $G$, we can put all $a$'s in the front and all $b$'s in the back, so we find that $$(ab)^{mn} = a^{mn}b^{mn} = (a^m)^n(b^m)^n = 1$$ where $1$ is the neutral element in your group $G$. This proves that $ab$ has finite order. Hence we have that $o(ab) = k$ for some non zero natural number $k$. From this, we find that $k \mid mn$.

Because we have that $(ab)^k = 1$ and $ab = ba$, we have that $a^k = b^{-k}$ and $b^k = a^{-k}$. Hence we also find the following equalities: $$(a^k)^n = (b^{-k})^n = (b^n)^{-k} = 1$$ and hence we must have that $m \mid kn$, since the order of $a$ is $m$. Analogously, we can show that we must have that $n \mid km$. Since $\text{gcd}(m,n) = 1$, we must have that $m \mid k$ and $n \mid k$, hence $mn \mid k$. Therefore, we have that $k = mn$, so $o(ab) = mn$.