# Prove the elements $(a,1)$ and $(1,b)$ of $A \times B$ commute and deduce that order of $(a,b)$ is lcm of order of a and b

I have no problem in commute part of the question. For lcm part

I did AS

Let $O(a)=m$ and $O(b)=n$ and let order of (a,b) be l. So $(a^l,b^l)=(1,1)$. So $a^l=1$ and $b^l=1$. Also i have $a^m=1$ and $b^n=1$. So i can say that $m,n$ divides $l$ (I am not sure about this as to whether $m|l$ or $l|m$ ). I need help from here

Another aproach i used is i take l=multiple of a and b. So $l=mk_1$ and $l=nk_2$.

CLAIM $(a,b)^l=(1,1)$

$a^{mk_1}=1$

$a^{nk_2}=1$ which is true because of order of a and b. So i am done with one part. Now to show that $l$ is the least mutiple is issue

To show $(a,1)$ commutes with $(1,b)$ you first need to define what multiplication means, but I assume by $(a_1,b_1)\star(a_2,b_2) = (a_1\cdot_A a_2, b_1\cdot_Bb_2)$ where $\cdot_A$ is the multiplication defined in the group $A$ and similar for $B$.

Now to show your two elements commute

$$(a, 1)\star(1, b) = (a\cdot_A 1, 1\cdot_B b) = (1\cdot_A a, b\cdot_B 1) = (1, b)\star (a,1)$$ Where we have used the fact that any element of your group commutes with the identity.

Now for the least common multiple part you need to multiply $a$ by itself $m$ times, and $b$ by itself $n$ times. If you do anything less than $\text{lcm}(a,b)$ then you will have some left overs $a$'s and $b$'s.

Let $$n=p_1^{\alpha_1}...p_r^{\alpha_r}q_1^{\beta_1}...q_1^{\beta_s}\\ m=p_1^{\alpha_1'}...p_r^{\alpha_r'}t_1^{\beta_1'}...t_k^{\beta_k'}$$ Here $gcd(p_{i_1},q_{j_1})=gcd(p_{i_2},t_{j_2})=gcd(q_{i_3},t_{j_3})=1$
for $1\leq p_{i_1},p_{i_2} \leq r$ and $1\leq q_{j_1},q_{i_3} \leq s$ and $1\leq t_{j_2},t_{j_3} \leq k$

$l$ must be a smallest integer such that $m|l$ and $n|l$ because of $l$ is order of $(a,b)$ .
For finding this we should choose $\gamma_i=max\{\alpha_i,\alpha_i'\}$ ,since $l$ is smallest integer such that $m|l$ and $n|l$ then
$$l=p_1^{\gamma_1}...p_r^{\gamma_r}q_1^{\beta_1}...q_1^{\beta_s}t_1^{\beta_1'}...t_k^{\beta_k'}=lcm(n,m)$$