# Use the second isomorphism theorem to conclude that $\gcd(a,b)\text{lcm}(a,b)=ab$

Use the second isomorphism theorem to conclude that $\gcd(a,b)\operatorname{lcm}(a,b)=ab$; that is, the product of the greatest common divisor and the lowest common multiple of $a,b$ is equal to $ab$.

I already know $(a) \cap (b) = (n)$, and $(a)+(b)=(d)$, where $n$ is lcm, $d$ is gcd.

then applying second second isomorphism theorem, we can know $(d)/(a)$ is isomorphic to $(b)/(n)$,

then, how to use this conclusion to show $dn=ab$, that is, $\gcd(a,b)\text{lcm}(a,b)=ab$?

Hint $\$ Comparing subgroup indices, $\,(d)/(a) = (b)/(n)\,\Rightarrow\,a/d = n/b\,$ $\Rightarrow$ $\,ab = dn$.

• Since it's isomorphism, both side have same cardinality, so divide everything by Z? – ZHJ Mar 9 '14 at 18:56
• Yes, compare cardinality. I don't know what you mean by "divide by $\Bbb Z$". – Bill Dubuque Mar 9 '14 at 19:03
• So comparing the cadinality is the key, but I am still a little confused about it. Could you write something explicitly about it? – ZHJ Mar 9 '14 at 19:10
• See for example the Wikipedia article index of a subgroup. – Bill Dubuque Mar 9 '14 at 19:25