The order of an element of a direct product of groups

[Mr Davis 97]

Homework Statement

Let $A$ and $B$ be finite groups, and $A \times B$ be their direct product. Given that $(a,1)$ and $(1,b)$ commute, and that $(a,1)^n = (a^n,1)$ and $(1,b)^n = (1,b^n)$ for all a and b, show that the order of $(a,b)$ is the least common multiple of the orders of a and b.

Homework Equations

The Attempt at a Solution

Let $n$ be the order of $(a,b)$. Then $(a,b)^n = (1,1)$,
$[(a,1)(1,b))]^n = (1,1)$
$(a,1)^n (1,b)^n = (1,1)$
$(a^n,1)(1,b^n) = (1,1)$
$(a^n,b^n) = (1,1)$
So $a^n = 1$and $b^n = 1$. It must be the case that n divides the order of $a$ and the order of $a$. The smallest number that divides both is by definition the least common multiple.

Is this proof okay?

 

[fresh_42]

Almost.

So $a^n = 1$ and $b^n = 1$.

O.k. until here.

It must be the case that n divides the order of $a$ and the order of $a$. The smallest number that divides both is by definition the least common multiple.

It must be the case that the order of $a$ and the order of $b$ divide $n$. The smallest number that is divided by both is by definition the least common multiple.

Do you know why $a^{ord(a)}=1=a^n$ implies $n\,\vert \,ord(a)\,$?

 

[Mr Davis 97]

Almost.

O.k. until here.

It must be the case that the order of $a$ and the order of $b$ divide $n$. The smallest number that is divided by both is by definition the least common multiple.

Do you know why $a^{ord(a)}=1=a^n$ implies $n\,\vert \,ord(a)\,$?

Right, my brain slipped up when I was typing. I know that if $a^n = 1$, then the order of a must divide $n$, specifically because the division algorithm

 

[Mr Davis 97]

Almost.

O.k. until here.

It must be the case that the order of $a$ and the order of $b$ divide $n$. The smallest number that is divided by both is by definition the least common multiple.

Do you know why $a^{ord(a)}=1=a^n$ implies $n\,\vert \,ord(a)\,$?

Right, my brain slipped up when I was typing. I know that if $a^n = 1$, then the order of a must divide $n$, specifically because the division algorithm

 

[nuuskur]

Indeed, you cannot have nonzero remainder, for it would contradict the minimality of the order, therefore $\mbox{ord}(a)\mid n$, I'm guessing fresh_42 meant it this way. The other way need not hold.

 

[fresh_42]

Right, my brain slipped up when I was typing.

Indeed, you cannot have nonzero remainder, for it would contradict the minimality of the order, therefore $\mbox{ord}(a)\mid n$, I'm guessing fresh_42 meant it this way. The other way need not hold.

Oops. Slippery ground, indeed.