# Proving something is commutative in abstract algebra

snesnerd
If $\ast : (f \ast g)(n) = \sum\limits_{d|n}f(d)g(\frac{n}{d})$, show that $\ast$ is commutative. Note that $d|n$ says $d$ divides $n$. Now I was not sure how to do this from an abstract algebra point of view although when I stare at it my though process was to maybe rewrite it somehow, which will then be easier to show its commutative.

## The Attempt at a Solution

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Homework Helper
If $\ast : (f \ast g)(n) = \sum\limits_{d|n}f(d)g(\frac{n}{d})$, show that $\ast$ is commutative. Note that $d|n$ says $d$ divides $n$. Now I was not sure how to do this from an abstract algebra point of view although when I stare at it my though process was to maybe rewrite it somehow, which will then be easier to show its commutative.

If $d|n$ then there exists a unique positive integer $c$ such that $cd = n$.

brmath
start by writing down g*f(n).

snesnerd
I understand what you mean pasmith, but I dont see how that would help show it is commutative.

Homework Helper
I understand what you mean pasmith, but I dont see how that would help show it is commutative.

Put, say n=6. Write out the sum f*g(6). Then write out g*f(6). Can you see why they are equal? That's what pasmith's hint is leading to.

snesnerd
Okay if that is the case, would the following proof show that it is commutative:

$(f \ast g)(n) = \sum\limits_{d|n}f(d)g(\frac{n}{d}) = \sum\limits_{d|n}f(\frac{n}{c})g(c) = \sum\limits_{d|n}g(c)f(\frac{n}{c}) = (g \ast f)(n)$

Homework Helper
Okay if that is the case, would the following proof show that it is commutative:

$(f \ast g)(n) = \sum\limits_{d|n}f(d)g(\frac{n}{d}) = \sum\limits_{d|n}f(\frac{n}{c})g(c) = \sum\limits_{d|n}g(c)f(\frac{n}{c}) = (g \ast f)(n)$

By itself, it doesn't make any sense. You have to work in the statement of how d is related to c and supply a few words of explanation.

snesnerd
$(f \ast g)(n) = \sum\limits_{d|n}f(d)g(\frac{n}{d})$.

Note that if $d|n$ then their exists a $c$ such that $n = cd$. Then $d = \frac{n}{c}$ and $c = \frac{n}{d}$. Substituting above gives us:

$\sum\limits_{d|n} f(\frac{n}{c})g(c)$

Now from here how would I word it to explain why $\sum\limits_{d|n}f(\frac{n}{c})g(c) = \sum\limits_{d|n}g(c)f(\frac{n}{c})$?

Homework Helper
$(f \ast g)(n) = \sum\limits_{d|n}f(d)g(\frac{n}{d})$.

Note that if $d|n$ then their exists a $c$ such that $n = cd$. Then $d = \frac{n}{c}$ and $c = \frac{n}{d}$. Substituting above gives us:

$\sum\limits_{d|n} f(\frac{n}{c})g(c)$

Now from here how would I word it to explain why $\sum\limits_{d|n}f(\frac{n}{c})g(c) = \sum\limits_{d|n}g(c)f(\frac{n}{c})$?

You want to explain why the set of divisors of n, and the set of numbers n/d where d is a divisor of n are exactly the same set of numbers.

snesnerd
Because if you consider the set of divisors of $n$ which are $cd$, and you substitute $\frac{n}{d}$ in for $c$ then you get that $n = (\frac{n}{d})d = n$.

Homework Helper
Because if you consider the set of divisors of $n$ which are $cd$, and you substitute $\frac{n}{d}$ in for $c$ then you get that $n = (\frac{n}{d})d = n$.

Ok, so if you sum f(d)*g(n/d) over all the divisors of n, then if you substitute the variable c=n/d (which also runs over all divisors of n) then you get the sum f(n/c)*g(c). There is a one-to-one correspondence between d and c.

snesnerd
But since we have that $n$ and $\frac{n}{d}$ are essentially the same set of numbers then then $\frac{n}{c}$ and $c$ are the same set of numbers so we can commute them and nothing will change/

But since we have that $n$ and $\frac{n}{d}$ are essentially the same set of numbers then then $\frac{n}{c}$ and $c$ are the same set of numbers so we can commute them and nothing will change/