Derive the identity elements of lcm and gcd

Question

Find the identity element of the binary operations $*,*'$ on $\mathbb{N}$ given by $a*b = lcm(a,b)$ and $a*'b = \gcd(a,b)$, where $\mathbb{N}=\{1,2,3...\}$

I know the identity element for lcm is $1$ and there is none for gcd in $\mathbb{N}$. But, how do I prove it mathematically ?

My Attempt

$a*e=a=e*a$

From the fundamental theorem of arithmetic, $$ a=\prod_ip_i^{a_i}\quad,e=\prod_ip_i^{e_i}\\ lcm(a,b)=\prod_ip_i^{\max(a_i,b_i)}\quad ,\gcd(a,b)=\prod_ip_i^{\min(a_i,b_i)} $$ For $*:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ $$ a*e=a\implies lcm(a,e)=\prod_ip_i^{\max(a_i,e_i)}=\prod_ip_i^{a_i}=a\\ \implies \max(a_i,e_i)=a_i\implies e_i\leq a_i \text{ for all }a_i\\ \implies e_i=0\implies \color{red}{e=1} $$ For $*':\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ $$ a*'e=a\implies \gcd(a,e)=\prod_ip_i^{\min(a_i,e_i)}=\prod_ip_i^{a_i}=a\\ \implies \min(a_i,e_i)=a_i\implies e_i\geq a_i\text{ for all }a_i\\ \implies \text{no }e_i\in\mathbb{N}\text{ satisfies the condition}\implies\color{red}{\text{no identity element for }*'} $$ Is it the right way to approach the problem ?

Answer 1

In your approach for $\gcd(a,e)=a$, you get that $\min(a_i,e_i)=a_i$ for all $i$. But keep in mind that $e$ will act as an identity for all elements $a$.

In particular, it should work as identity if we consider $a=e^2$, i.e. we should have $\gcd(e^2,e)=e^2$. But the prime factorization of $e^2$ will of the form $\prod_ip_i^{2e_i}$. So here we have $a_i=2e_i > e_i$. But this violates $\min(2e_i,e_i)=2e_i$. So no such $e$ exists.

Note: But as suggested by many others, going through prime factorization approach is an overkill for this problem.

Answer 2

$a*b = lcm(a,b)$

Prove that if $lcm(a,e) = a$ for all natural $a$ then $e = 1$.

Pf: $lcm(a,e) = \frac {ae}{\gcd(a,e)}$.

So if $lcm(a,e) = \frac {ae}{\gcd(a,e)} =a$ for all natural $a$ then

$ae = a*\gcd(a,e)$ for all natural numbers.

So $e = \gcd(a,e)$ for all natural numbers.

So $e$ divides all natural numbers.

So $e$ divides $1$. The only natural number that divides $1$ is $1$.

So $e = 1$.

And indeed $a*1 = lcm(a,1) = a$ for all $a$.

$a *' b = \gcd(a,e)$

So prove find an $e$ so that $\gcd(a,e) = a$ for all $a$ or prove one can't exist.

S'pose there were such an $e$ then $\gcd(a,e) =a$ for all possible $a$.

So all $a$ will divide $e$. So we must find a number that is divisible by any other number.

Well, that would mean $2e|e$ so there is an integer $k$ so that $e = k*(2e)$ so if $e\ne 0$ then $k = \frac 12$ which is not an integer. So $e = 0$.

And .... every number does divide $0$ because $a*0 = 0$ for every number.

And $\gcd(a,0) = a$. Because $a|0$ and $a|a$. And if $m > a > 0$ then $m\not \mid a$.

But the question now is, was it specified that the identity element must be in $\mathbb N$?

I guess so because otherwize $-1$ would be an identity element for $*$. SO as $0 \not \in \mathbb N$ I guess we can say there is no natural number that is an identity.